Polynomials and Equations

Problems

Solutions

Solution: PRMO 2012, Q1

Let the number be $N$. Rachel computed $(N - 9)/3 = 43$, so $N - 9 = 129$ and $N = 138$.

The intended calculation was $$\frac{138 - 3}{9} = \frac{135}{9} = 15.$$

Answer 15

Solution: PRMO 2013, Q3

If the roots are integers $r$ and $s$ then $rs = 20$ and $a = -(r+s)$. The factor pairs of $20$ over the integers come in two families, positive and negative: $$(1,20), (2,10), (4,5) \quad \text{and} \quad (-1,-20), (-2,-10), (-4,-5),$$ giving $a = -21, -12, -9$ and $a = 21, 12, 9$ respectively.

Every value is matched by its negative, because negating both roots leaves the product unchanged and reverses the sum. So the six values cancel in pairs and their total is $$0.$$

Answer 0

Solution: PRMO 2015 Part A, Q1

Let $w$ be the time for one walk of that distance and $r$ the time for one ride. The two facts we are given translate directly: $$w + r = 3\tfrac34 = \tfrac{15}{4}, \qquad 2r = 2\tfrac12 = \tfrac52.$$ The second gives $r = \tfrac54$ at once, and then the first gives $$w = \tfrac{15}{4} - \tfrac54 = \tfrac{10}{4} = \tfrac52.$$ Walking both ways therefore takes $2w = 5$ hours.

The reason the arithmetic is so short is that the second condition was handed to us already solved for a single ride. A question phrased as three separate journey times often hides one equation that needs no work at all, and it pays to look for it before setting up anything larger.

Answer 5

Solution: PRMO 2015 Part A, Q15

Adding the two equations gives $2 \cdot 3^x = 1458$, so $$3^x = 729 = 3^6, \qquad x = 6.$$ Subtracting them gives $2 \cdot 2^y = 512$, so $$2^y = 256 = 2^8, \qquad y = 8.$$ Therefore $xy = 48$.

The only thing to be careful about is recognising $729$ and $256$ as exact powers rather than reaching for logarithms, which the competition never intends. Building up the small powers of $2$ and $3$ from memory is worth the five minutes it takes.

Answer 48

Solution: PRMO 2017, Q3

Team $A$ does $\tfrac1{12}$ of the job each day and team $B$ does $\tfrac1{36}$.

In the first four days, $A$ alone completes $\tfrac{4}{12} = \tfrac13$. In the next two days both teams work, completing $$2\left(\frac1{12} + \frac1{36}\right) = 2 \cdot \frac{3 + 1}{36} = \frac{8}{36} = \frac29.$$ So after six days the fraction done is $\tfrac13 + \tfrac29 = \tfrac59$, leaving $\tfrac49$.

Team $B$ alone needs $$\frac{4/9}{1/36} = \frac49 \times 36 = 16$$ more days.

Answer 16

Solution: IOQM 2025 Part SEP, Q7

Both quantities in the question are $\dfrac{60x}{100}$: the first reads "$60$ percent of $x$" and the second "$x$ percent of $60$". Since we are told the first equals $40$, so does the second.

If you would rather see the number, $60\%$ of $x$ is $40$ gives $x = \tfrac{200}{3}$, and then $x\%$ of $60$ is $\tfrac{200}{3} \cdot \tfrac{60}{100} = 40$, as promised.

Answer 40

Solution: PRMO 2013, Q11

Add the three equations: $$x^2 + y^2 + z^2 + 2x + 6y + 4z = -17 + 1 + 2 = -14.$$ Move everything to one side and complete each square: $$(x+1)^2 + (y+3)^2 + (z+2)^2 = -14 + 1 + 9 + 4 = 0.$$

A sum of three real squares vanishes only when each does, so $$x = -1, \qquad y = -3, \qquad z = -2,$$ and these do satisfy all three original equations. Hence $$x^2 + y^2 + z^2 = 1 + 9 + 4 = 14.$$

Answer 14

Solution: PRMO 2014, Q5

Call the common value $t$. Reading the chain from the left, $$a = t - 1, \quad b = t - 2, \quad c = t - 3, \quad d = t - 4, \quad e = t - 5,$$ so their sum is $5t - 15$. The final expression in the chain says $$t = (a+b+c+d+e) + 3 = 5t - 15 + 3 = 5t - 12,$$ hence $4t = 12$ and $t = 3$.

The five numbers are therefore $2, 1, 0, -1, -2$, and $$a^2 + b^2 + c^2 + d^2 + e^2 = 4 + 1 + 0 + 1 + 4 = 10.$$

Answer 10

Solution: PRMO 2014, Q6

If the roots are integers $r$ and $s$ then $rs = 2014$ and $r + s = n$. Since $2014 > 0$ and $n$ is a natural number, both roots are positive.

Factorising, $2014 = 2 \times 19 \times 53$, so the factor pairs and their sums are $$(1, 2014) \to 2015, \qquad (2, 1007) \to 1009,$$ $$(19, 106) \to 125, \qquad (38, 53) \to 91.$$ The sum is smallest for the most balanced pair, which is the general rule for a fixed product, so $$n = 91.$$

Answer 91

Solution: PRMO 2014, Q7

Work with the size of $x$ rather than its sign, which settles both cases at once. Write $t = |x|$. Whatever $x$ is, $x^4 = t^4$, and if the power $x^{(x^4)}$ is a real number then its size is $|x|$ raised to that same exponent. So the equation forces $$t^{\,t^4} = 4.$$

That has exactly one positive solution. For $0 < t \le 1$ the left side is at most $1$, so $t > 1$; and for $t > 1$ both $t$ and $t^4$ increase with $t$, so $t^{\,t^4}$ is strictly increasing and meets the value $4$ once. Since $\left(\sqrt2\right)^4 = 4$, that once is at $$t = \sqrt2.$$

So $x = \sqrt2$ or $x = -\sqrt2$, and both really do work. The first is immediate. For the second, the exponent $x^4$ is $4$, a whole number, so the power is defined without any question about what a negative base raised to a fractional power should mean, and $\left(-\sqrt2\right)^4 = 4$ as well.

Both signs give the same answer, because the two exponents in question are also even. With $x = \pm 2^{1/2}$, $$x^{(x^2)} = 2^{\frac12 \times 2} = 2, \qquad x^{(x^8)} = 2^{\frac12 \times 16} = 2^8 = 256.$$ Their sum is $$2 + 256 = 258.$$

The exponents $x^2$, $x^4$, $x^8$ are successively squared, so the three values are $2^1$, $2^2$ and $2^8$: the answer grows alarmingly fast, which is the joke in the question.

Answer 258

Solution: PRMO 2014, Q8

Let $S$ have $n$ elements and total $nM$.

First note that neither $15$ nor $1$ already lies in $S$, which is what makes the two unions bigger. If $15$ were in $S$ then $S \cup \{15\}$ would be $S$ itself, with mean $M$ rather than $M+2$. And if $1$ were in $S$ then $S \cup \{15, 1\}$ would be $S \cup \{15\}$, with mean $M+2$ rather than $M+1$. So the two unions have $n+1$ and $n+2$ elements.

Adding $15$ gives a set of $n+1$ elements with mean $M + 2$: $$nM + 15 = (n+1)(M+2) = nM + M + 2n + 2,$$ so $M + 2n = 13$.

Adding $15$ and $1$ gives $n + 2$ elements with mean $M + 1$: $$nM + 16 = (n+2)(M+1) = nM + 2M + n + 2,$$ so $2M + n = 14$.

Solving the pair, $M = 13 - 2n$ and $2(13 - 2n) + n = 14$, so $3n = 12$ and $$n = 4, \qquad M = 5.$$ A check: a four-element set of mean $5$ has total $20$; adding $15$ gives $35$ over five elements, mean $7 = M+2$; adding $1$ as well gives $36$ over six, mean $6 = M+1$.

Answer 4

Solution: PRMO 2015 Part A, Q3

Suppose $r$ satisfies both $x^2 - 4x + k = 0$ and $x^2 + kx - 4 = 0$. Subtracting the second from the first removes the quadratic term: $$(-4 - k)r + (k + 4) = 0, \qquad \text{that is} \qquad (k+4)(1 - r) = 0.$$ So either $k = -4$ or $r = 1$, and the two cases are worth separating carefully.

If $k = -4$ the two equations both read $x^2 - 4x - 4 = 0$. They are then the same equation and share both roots, which the problem has ruled out by insisting on exactly one common root.

That leaves $r = 1$. Putting $x = 1$ into the first equation gives $1 - 4 + k = 0$, so $k = 3$. It is worth confirming that this really does produce exactly one shared root and not two: with $k = 3$ the equations factorise as $(x-1)(x-3) = 0$ and $(x-1)(x+4) = 0$, whose roots are $\{1,3\}$ and $\{1,-4\}$. They meet in $1$ alone, as required.

Answer 3

Solution: PRMO 2017, Q4

If all roots are negative integers, write the roots of the first quadratic as $-p$ and $-q$ with $p, q$ positive integers, and those of the second as $-r$ and $-s$. Comparing coefficients, $$pq = 20, \quad a = p + q, \qquad r + s = 17, \quad b = rs.$$ The two quadratics share no constraint, so $a$ and $b$ can be minimised separately.

For $a$: the factor pairs of $20$ are $(1,20)$, $(2,10)$ and $(4,5)$, giving $a = 21, 12, 9$. The smallest is $a = 9$, which is the general rule that a product splits most evenly nearest its square root.

For $b$, use positivity of the integers $r,s$ and their sum $17$: $$rs-16=rs-(r+s)+1=(r-1)(s-1)\ge0.$$ Equality is attained at $(r,s)=(1,16)$, so the least $b$ is $16$.

Hence the smallest value of $a + b$ is $$9 + 16 = 25.$$

Note that the two halves pull in opposite directions: a fixed product wants a balanced split, a fixed sum wants a lopsided one. Both are the same fact about $(r-s)^2 \ge 0$ seen from different sides.

Answer 25

Solution: PRMO 2017, Q19

Write the equation as $x^4 + 0 \cdot x^3 + ax^2 + bx - c = 0$, a monic quartic with four roots, three of which are $1$, $2$ and $3$. Let the fourth be $r$: the factor theorem gives the three factors $x-1,x-2,x-3$, and the remaining monic factor is linear, $x-r$. In $$(x-1)(x-2)(x-3)(x-r),$$ the $x^3$ term takes a constant from one factor and $x$ from the other three, so its coefficient is $-(1+2+3+r)$. The constant term takes a constant from every factor, giving $1\cdot2\cdot3\cdot r$.

The coefficient of $x^3$ is zero, and for a monic quartic that coefficient is minus the sum of the roots. So $$1 + 2 + 3 + r = 0, \qquad r = -6.$$

The constant term of a monic quartic is the product of its roots. Here the constant term is $-c$, so $$-c = 1 \cdot 2 \cdot 3 \cdot (-6) = -36, \qquad c = 36.$$

Neither $a$ nor $b$ was needed, though they follow at once: the polynomial is $(x-1)(x-2)(x-3)(x+6)$.

Answer 36

Solution: PRMO 2018, Q1

Let the first volume have $x$ pages. Then the second has $x + 50$ and the third has $\tfrac32(x+50)$.

The first page of volume one is numbered $1$. The first page of volume two follows the whole of volume one, so it is $x + 1$. The first page of volume three follows volumes one and two, so it is $x + (x+50) + 1 = 2x + 51$. Adding, $$1 + (x+1) + (2x+51) = 3x + 53 = 1709,$$ so $3x = 1656$ and $x = 552$. The three volumes then have $552$, $602$ and $903$ pages, and the last page number is $$n = 552 + 602 + 903 = 2057.$$

Factorising, $2057 = 11 \times 187 = 11 \times 11 \times 17$, so the largest prime factor is $17$.

The condition that the third volume is one and a half times the second is what forces $x + 50$ to be even, and $602$ duly is. A problem that hands you a fractional multiplier is usually checking that you notice the parity it demands.

Answer 17

Solution: PRMO 2019, Q2

Write $f(x) = x^2 + ax + b$ and expand the two sides for nonzero $x$. On the left, $$f\!\left(x + \tfrac1x\right) = \left(x + \tfrac1x\right)^2 + a\left(x + \tfrac1x\right) + b,$$ which expands to $x^2 + \tfrac{1}{x^2} + 2 + a\left(x + \tfrac1x\right) + b$. On the right, $$f(x) + f\!\left(\tfrac1x\right) = x^2 + ax + b + \tfrac{1}{x^2} + \tfrac{a}{x} + b = x^2 + \tfrac{1}{x^2} + a\left(x + \tfrac1x\right) + 2b.$$ Everything cancels except the constants, leaving $2 + b = 2b$, that is $b = 2$. Notice that $a$ is untouched: the functional equation constrains only the constant term.

Now use the second condition. The roots of $x^2 + ax + 2 = 0$ are integers with product $2$, so they are $1$ and $2$, or $-1$ and $-2$. Their sum is $-a$, giving $a = -3$ or $a = 3$. Either way $a^2 = 9$, and $$a^2 + b^2 = 9 + 4 = 13.$$

Answer 13

Solution: IOQM 2023, Q2

The two logarithms in the equation are reciprocals of one another, since $\log_b a = 1/\log_a b$. Writing $t = \log_a b$, which is legitimate because $a, b \ge 2$, the condition becomes $$t + \frac{6}{t} = 5, \qquad \text{that is} \qquad t^2 - 5t + 6 = 0,$$ so $t = 2$ or $t = 3$. In other words $b = a^2$ or $b = a^3$, and nothing else can occur.

Counting is now a matter of respecting the bound $b \le 2023$.

For $b = a^2$ we need $a^2 \le 2023$, and since $44^2 = 1936$ while $45^2 = 2025$, the values $a = 2, 3, \ldots, 44$ all work. That is $43$ pairs.

For $b = a^3$ we need $a^3 \le 2023$, and since $12^3 = 1728$ while $13^3 = 2197$, the values $a = 2, 3, \ldots, 12$ work, giving $11$ pairs.

No pair is counted twice, because $a^2 = a^3$ only for $a = 1$, which is excluded. The total is $43 + 11 = 54$.

Answer 54

Solution: IOQM 2024, Q5

Set $$p = \frac xy, \qquad q = \frac yz, \qquad r = \frac zx, \qquad \text{so that } pqr = 1.$$ Then $a = p + q + r$ straight away. For $b$, note that $\dfrac xz = \dfrac1r$ and $\dfrac yx = \dfrac1p$ and $\dfrac zy = \dfrac1q$, so $$b = \frac1p + \frac1q + \frac1r = \frac{qr + rp + pq}{pqr} = pq + qr + rp,$$ using $pqr = 1$. So $a$ and $b$ are the first two elementary symmetric functions of $p, q, r$, and the third is $pqr = 1$.

Now $c = (p+q)(q+r)(r+p)$, and there is a standard identity for that product: $$(p+q)(q+r)(r+p) = (p+q+r)(pq+qr+rp) - pqr.$$ It is worth seeing why rather than taking it on trust. Write out the two expansions: $$\begin{aligned} (p+q)(q+r)(r+p)&=p^2q+p^2r+pq^2+pr^2+q^2r+qr^2+2pqr,\\ (p+q+r)(pq+qr+rp)&=p^2q+p^2r+pq^2+pr^2+q^2r+qr^2+3pqr. \end{aligned}$$ The six mixed-square terms match; the second line has one extra $pqr$.

Hence $c = ab - 1$ and $$|ab - c| = |ab - (ab-1)| = 1.$$

Answer 1

Solution: IOQM 2024, Q16

The relation ties $f(x)$ to $f(3-x)$, so apply it at $3-x$ as well, noting that $3 - (3-x) = x$: $$4f(3-x) + 3f(x) = x^2, \qquad 4f(x) + 3f(3-x) = (3-x)^2.$$ Treat $f(x)$ and $f(3-x)$ as two unknowns. Multiplying the first by $3$, the second by $4$, and subtracting removes $f(3-x)$: $$7f(x) = 4(3-x)^2 - 3x^2 = 36 - 24x + x^2,$$ so $$f(x) = \frac{x^2 - 24x + 36}{7}.$$

Now the required difference is exact, with no rounding needed: $$f(27) - f(25) = \frac{(729 - 648 + 36) - (625 - 600 + 36)}{7} = \frac{56}{7} = 8.$$

Answer 8

Solution: IOQM 2025 Part SEP, Q28

First exclude $n=1$: it gives $f(1)=0$, outside the domain of $f$, so $f(f(1))$ is undefined. For $n\ge2$, take the two parity cases.

If $n$ is odd then $f(n) = n-1$, which is even, so $$f(f(n)) = (n-1)^2 - 1 = 99, \qquad (n-1)^2 = 100, \qquad n = 11,$$ taking the positive root since $n$ is a positive integer. And $11$ is indeed odd, so this is a genuine solution.

If $n$ is even then $f(n) = n^2-1$, which is odd, so $$f(f(n)) = (n^2-1) - 1 = n^2 - 2 = 99, \qquad n^2 = 101,$$ which is not a perfect square. No solution here.

So the only value is $n = 11$, and the sum of all possible values is $11$.

Answer 11

Solution: PRMO 2012, Q12

To remove the nested root, try a sum $\sqrt u+\sqrt v$. Its square is $u+v+2\sqrt{uv}$, so we want $$u+v=2011,\qquad4uv=2011^2-1.$$ Then $(u-v)^2=(u+v)^2-4uv=1$, giving $u=1006$ and $v=1005$. These numbers suggest the following identity, which also checks the calculation: $$\left(\frac{\sqrt{a+1} + \sqrt{a-1}}{\sqrt2}\right)^2 = \frac{2a + 2\sqrt{a^2-1}}{2} = a + \sqrt{a^2 - 1}.$$

With $a = 2011$ that gives $$\frac{1}{\sqrt{2011 + \sqrt{2011^2-1}}} = \frac{\sqrt2}{\sqrt{2012} + \sqrt{2010}}.$$ Rationalising the denominator, whose conjugate difference is $2012 - 2010 = 2$, $$\frac{\sqrt2\left(\sqrt{2012} - \sqrt{2010}\right)}{2} = \frac{\sqrt{4024} - \sqrt{4020}}{2} = \sqrt{1006} - \sqrt{1005}.$$

So $m = 1006$ and $n = 1005$ is one answer, and $m + n = 2011$.

Why those two integers are forced

Nothing so far rules out a second pair, and the question asks for $m+n$ as though there were only one, so it is worth checking. Suppose the same number equals $\sqrt m - \sqrt n$ for positive integers $m$ and $n$, and put $k = m - n$, which is positive since $\sqrt m > \sqrt n$. Because $$\left(\sqrt{1006} - \sqrt{1005}\right)\left(\sqrt{1006} + \sqrt{1005}\right) = 1,$$ the number is the reciprocal of $\sqrt{1006} + \sqrt{1005}$, and likewise $\left(\sqrt m - \sqrt n\right)\left(\sqrt m + \sqrt n\right) = k$ gives $\sqrt m + \sqrt n = k\left(\sqrt{1006} + \sqrt{1005}\right)$. Adding this to $\sqrt m - \sqrt n = \sqrt{1006} - \sqrt{1005}$, $$2\sqrt m = (k+1)\sqrt{1006} + (k-1)\sqrt{1005}.$$ Squaring, the cross term is $2(k+1)(k-1)\sqrt{1006 \cdot 1005}$, and $1006 \cdot 1005 = 1011030$ lies strictly between $1005^2$ and $1006^2$, so it is not a perfect square and its root is irrational. Everything else in the equation is a whole number, so $(k+1)(k-1) = 0$, and with $k$ positive that means $k = 1$. Then $2\sqrt m = 2\sqrt{1006}$, so $m = 1006$ and $n = 1005$ after all, and the answer stands.

The answer being the year is not a coincidence: for any $a$ the same computation gives $\sqrt{\tfrac{a+1}{2}} - \sqrt{\tfrac{a-1}{2}}$, whose two arguments add to $a$.

Answer 2011

Solution: PRMO 2012, Q13

Add the three relations: $$a + b + c = (b - c) + (c - d) + (d - a) = b - a,$$ so $2a + c = 0$, that is $c = -2a$. Then $c = d - a$ gives $d = c + a = -a$, and $b = c - d$ gives $b = -2a + a = -a$. The first relation is then satisfied automatically: $b - c = -a + 2a = a$.

Since $abcd \ne 0$ we have $a \ne 0$, and $$\frac ab + \frac bc + \frac cd + \frac da = \frac{a}{-a} + \frac{-a}{-2a} + \frac{-2a}{-a} + \frac{-a}{a} = -1 + \frac12 + 2 - 1 = \frac12.$$

Answer $1/2$

Solution: PRMO 2012, Q17

Write $P(x) = x^3 + 3x + 5 = (x - x_1)(x - x_2)(x - x_3)$. Since $x_i + 1/x_i = \left(x_i^2+1\right)/x_i$, $$\prod_{i=1}^{3}\left(x_i + \frac{1}{x_i}\right) = \frac{\prod_i \left(x_i^2 + 1\right)}{\prod_i x_i}.$$

The denominator is Vieta on $x^3 + 0x^2 + 3x + 5$, which gives $x_1x_2x_3 = -5$.

For the numerator, reflect. Because $P$ has odd degree, $$P(-x) = \prod_i \left(-x - x_i\right) = -\prod_i \left(x + x_i\right),$$ so $\prod_i \left(x + x_i\right) = -P(-x)$, and multiplying the two products pairs each root with its negative: $$\prod_i \left(x^2 - x_i^2\right) = P(x)\left[-P(-x)\right] = \left(x^3 + 3x + 5\right)\left(x^3 + 3x - 5\right),$$ a difference of two squares. So $$\prod_i \left(x^2 - x_i^2\right) = \left(x^3 + 3x\right)^2 - 25 = x^2\left(x^2+3\right)^2 - 25.$$ Both sides are polynomials in $x^2$ that agree for every real $x$, so writing $y = x^2$ they agree for every $y \ge 0$, and two polynomials that agree at infinitely many points are the same polynomial: $$\prod_i \left(y - x_i^2\right) = y(y+3)^2 - 25.$$

Now put $y = -1$, which is legitimate because that is an identity between polynomials rather than a statement about a square. The left side becomes $\prod_i\left(-1 - x_i^2\right) = -\prod_i\left(x_i^2+1\right)$, and the right side is $(-1)(-1+3)^2 - 25 = -29$. So the numerator is $29$, and $$\prod_{i=1}^{3}\left(x_i + \frac{1}{x_i}\right) = \frac{29}{-5} = -\frac{29}{5}.$$

Answer $-29/5$

Solution: PRMO 2014, Q9

Let $a$ and $b$ be the roots of the first quadratic, so $a + b = k$ and $ab = l$, both natural numbers. In particular $ab = l \ge 1$, so neither root is zero and the reciprocals below are defined. The new quadratic has roots $a + \tfrac1b$ and $b + \tfrac1a$, so its coefficients are

$$p = a + b + \frac1a + \frac1b = k + \frac{a+b}{ab} = k + \frac{k}{l},$$ $$q = \left(a + \frac1b\right)\left(b + \frac1a\right) = ab + 1 + 1 + \frac{1}{ab} = l + 2 + \frac1l.$$

Now $q$ is required to be a natural number, and $l$ is one, so $\tfrac1l$ must be an integer: $l = 1$. Then $$q = 1 + 2 + 1 = 4,$$ and $p = 2k$, which is a natural number for every natural $k$, so such $k, l, p, q$ really do exist.

The value of $q$ is therefore forced, and the sum of all possible values is $4$.

Answer 4

Solution: PRMO 2015 Part A, Q12

Two equations in three unknowns cannot determine $a$, $b$ and $c$ separately, so the quantity asked for must be constant along the whole line of solutions. Let us find that line by solving for $a$ and $c$ in terms of $b$.

Multiply the second equation $8a + 4b - c = 7$ by $8$ and add the first, $a - 7b + 8c = 4$: $$65a + 25b = 60, \qquad \text{so} \qquad 13a + 5b = 12.$$ Substituting $a = \tfrac{12 - 5b}{13}$ back into the first equation and clearing denominators gives $8c = 4 - a + 7b = \tfrac{40 + 96b}{13}$, hence $$13a = 12 - 5b, \qquad 13c = 5 + 12b.$$

Now square and add. The cross terms are $-120b$ and $+120b$, which cancel: $$169(a^2 + c^2) = (12 - 5b)^2 + (5 + 12b)^2 = 169 + 169b^2 = 169(1 + b^2).$$ Dividing by $169$ gives $a^2 + c^2 = 1 + b^2$, and therefore $$a^2 - b^2 + c^2 = 1.$$

The cancellation was not luck. The coefficient pairs $(12,-5)$ and $(5,12)$ are perpendicular and both have length $13$, which is exactly the condition under which the sum of the two squares loses its cross term and keeps a clean multiple of $1 + b^2$. Ugly-looking coefficients in a problem like this are usually built to do something of that kind.

Answer 1

Solution: PRMO 2015 Part A, Q18

Start with the denominators, which look unpromising until the condition is used. Since $c = -a-b$, $$2a^2 + bc = 2a^2 + b(-a-b) = 2a^2 - ab - b^2 = (2a+b)(a-b),$$ and $2a + b = a + (a+b) = a - c$. So $$2a^2 + bc = (a-b)(a-c),$$ and by the same computation with the letters rotated, $2b^2 + ca = (b-c)(b-a)$ and $2c^2 + ab = (c-a)(c-b)$. That $P$ is defined tells us these are all non-zero, so $a$, $b$, $c$ are three distinct numbers. What we must evaluate is $$P = \frac{a^2}{(a-b)(a-c)} + \frac{b^2}{(b-a)(b-c)} + \frac{c^2}{(c-a)(c-b)}.$$

At this point the condition $a + b + c = 0$ has done its job and is no longer needed, because the sum above equals $1$ for any three distinct numbers. Here is the reason, and the one fact behind it is that a quadratic is completely determined by its values at three different points: two quadratics that agree at three points have a difference of degree at most $2$ with three roots, so that difference is the zero polynomial.

To make one term select the value at $a$, it should vanish at $b$ and $c$, so give it the factors $(x-b)(x-c)$; divide by $(a-b)(a-c)$ to make its value at $a$ equal $1$. For a concrete example at the points $0,1,2$, the polynomial $-x(x-2)$ is $1$ at $1$ and $0$ at the other two. Apply this same recipe at $a,b,c$ and consider $$L(x) = a^2 \frac{(x-b)(x-c)}{(a-b)(a-c)} + b^2 \frac{(x-a)(x-c)}{(b-a)(b-c)} + c^2 \frac{(x-a)(x-b)}{(c-a)(c-b)},$$ which has degree at most $2$. Each of the three fractions is designed to equal $1$ at one of the three points and $0$ at the other two, so substituting $x = a$ makes the last two terms vanish and the first collapse to $a^2$, and similarly at $x = b$ and $x = c$. So $L$ agrees with $x^2$ at $a$, $b$ and $c$, and by the fact just stated $L(x) = x^2$ identically.

Comparing the coefficients of $x^2$ on both sides gives exactly $P = 1$, since the coefficient of $x^2$ in $L$ is the sum $P$ and the coefficient of $x^2$ in $x^2$ is $1$.

That way of assembling a polynomial from its values is called Lagrange interpolation. Here it explains why the sum is the coefficient of $x^2$, rather than merely confirming its value.

Answer 1

Solution: PRMO 2017, Q2

Put $x = \sqrt a$ and $y = \sqrt b$, both positive. The two conditions become $$x^3 + y^3 = 183, \qquad x^2 y + x y^2 = xy(x+y) = 182.$$

Now recall the expansion $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$, which is built from precisely those two quantities: $$(x+y)^3 = 183 + 3 \times 182 = 183 + 546 = 729, \qquad \text{so} \qquad x + y = 9.$$

From the second condition, $9xy = 182$, so $xy = \tfrac{182}{9}$. Hence $$a + b = x^2 + y^2 = (x+y)^2 - 2xy = 81 - \frac{364}{9} = \frac{729 - 364}{9} = \frac{365}{9},$$ and $$\frac95 (a+b) = \frac95 \cdot \frac{365}{9} = \frac{365}{5} = 73.$$

The factor $\tfrac95$ in the question is a courtesy: it turns an ugly fraction into a whole number, and its presence is a hint that $a+b$ will have a $9$ underneath.

Answer 73

Solution: PRMO 2018, Q9

Write $s = a + b$, so $b = s - a$, and substitute $x = s$ into the equation: $$s^2 + as + b = 0, \qquad \text{that is} \qquad s^2 + as + s - a = 0.$$ Collecting the terms in $a$ gives $a(s - 1) = -(s^2 + s)$, and $s = 1$ is impossible because it would force $s^2 + s = 2$ to vanish. So $$a = -\frac{s^2 + s}{s - 1} = -(s+2) - \frac{2}{s-1},$$ using $s^2 + s = (s-1)(s+2) + 2$.

For $a$ to be an integer, $s - 1$ must divide $2$, so $s \in \{-1, 0, 2, 3\}$. Working out $a$ and then $b = s - a$ in each case: $$\begin{array}{c|rrr} s & a & b & b^2 \\ \hline -1 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 2 & -6 & 8 & 64 \\ 3 & -6 & 9 & 81 \end{array}$$ Each really works: for instance $s = 3$ gives $x^2 - 6x + 9 = (x-3)^2$, whose root $3$ is indeed $a + b$. So the maximum of $b^2$ is $81$.

Answer 81

Solution: PRMO 2019, Q8

Everything turns on one factorisation: $x^3 - 1 = (x-1)\left(x^2+x+1\right)$, so $x^2+x+1$ divides $x^3 - 1$. It then divides $x^{3m}-1$ for every $m \ge 1$ as well, since $$x^{3m} - 1 = \left(x^3 - 1\right)\left(x^{3m-3} + x^{3m-6} + \cdots + 1\right).$$

Write $2^n = 3m + r$ with remainder $r$, so $0 \le r \le 2$. Then $$x^{2^n} - x^{r} = x^{r}\left(x^{3m} - 1\right)$$ is divisible by $x^2+x+1$, which says that $x^{2^n} + x + 1$ and $x^{r} + x + 1$ leave the same remainder on division by $x^2+x+1$. So one is divisible by it exactly when the other is, and since $2^n$ is never a multiple of $3$ there are only two cases to try: $$r = 1: \quad x + x + 1 = 2x+1, \qquad\qquad r = 2: \quad x^2 + x + 1.$$ A polynomial of degree $1$ is not divisible by one of degree $2$, so $r=1$ fails; $r=2$ gives the divisor itself, so it succeeds. The condition is exactly $$2^n \equiv 2 \pmod 3.$$ Now $2 \equiv -1 \pmod 3$, so $2^n \equiv (-1)^n$, which is $2 \equiv -1$ exactly when $n$ is odd.

The odd $n$ with $3 \le n \le 100$ are $3, 5, 7, \ldots, 99$, and there are $\tfrac{99-3}{2} + 1 = 49$ of them.

Answer 49

Solution: IOQM 2023, Q12

The relevant identity is $$x^3+y^3+z^3-3xyz = (x+y+z)\cdot\tfrac12\bigl[(x-y)^2+(y-z)^2+(z-x)^2\bigr],$$ so the given condition holds exactly when $p_1+p_2+p_3 = 0$ or $p_1 = p_2 = p_3$.

Look at the first possibility. Adding the three values, $$p_1+p_2+p_3 = (1+8+27) + (1+4+9)a + (1+2+3)b + 3c,$$ that is, $36 + 14a + 6b + 3c$. Every term except $3c$ is even, and $c$ is odd, so $3c$ is odd and the whole sum is odd. An odd number is not $0$. So this possibility cannot occur, and the parity condition on $c$, which looks decorative, is exactly what removes it.

That leaves $p_1 = p_2 = p_3$, say all equal to $k$. Then $P(x) - k$ is a monic cubic vanishing at $1$, $2$ and $3$, so $$P(x) = (x-1)(x-2)(x-3) + k.$$ Its constant term is $c = P(0) = -6 + k$, which is odd precisely when $k$ is odd, so such polynomials do exist. Reading off the values we need, $$p_0 = k - 6, \qquad p_1 = p_2 = k,$$ and therefore $$p_2 + 2p_1 - 3p_0 = k + 2k - 3(k-6) = 18.$$

The answer does not depend on $k$, which is the sign that the question was built around the factorisation rather than around any particular polynomial.

Answer 18

Solution: IOQM 2024, Q10

Group the terms so that the squares appear: $$a^2 - 6ab + 9b^2 = (a - 3b)^2, \qquad p^2b^2 - 6pbc + 9c^2 = (pb - 3c)^2,$$ and adding these accounts for every term of the given expression, since $9b^2 + p^2b^2 = (p^2+9)b^2$. So the equation reads $$(a-3b)^2 + (pb - 3c)^2 = 0.$$ A sum of two real squares vanishes only when both vanish, so $$a = 3b, \qquad c = \frac{pb}{3}.$$

Now impose the triangle inequalities on the sides $3b$, $b$, $\tfrac{pb}{3}$, and divide throughout by the positive number $b$: $$1 + \frac p3 > 3 \ \Rightarrow\ p > 6, \qquad 3 + 1 > \frac p3 \ \Rightarrow\ p < 12,$$ while the third inequality $3 + \tfrac p3 > 1$ holds automatically.

So $p \in \{7, 8, 9, 10, 11\}$, and every one of these does produce a genuine triangle: take $b = 3$, so that the sides are $$a = 9, \qquad b = 3, \qquad c = p,$$ and the two binding inequalities read $3 + p > 9$ and $9 + 3 > p$, both of which hold for each of the five values. There are $5$ of them.

Answer 5

Solution: IOQM 2025 Part SEP, Q7

Set $n = 0$ in the relation. Since $mn + 1 = 1$, $$f(1) = f(m)f(0) - f(0) - m + 2 = f(m) - m + 1,$$ using $f(0) = 1$. Rearranged, $f(m) = f(1) + m - 1$ for every integer $m$, so $f$ is linear.

Now find the constant. Taking $m = 0$ in that formula gives $f(0) = f(1) - 1$, and $f(0) = 1$, so $f(1) = 2$ and $$f(m) = m + 1.$$ It is worth confirming this really satisfies the original relation, since we only used one special case of it: $$f(mn+1) = mn + 2,$$ $$f(m)f(n) - f(n) - m + 2 = (m+1)(n+1) - (n+1) - m + 2 = mn + 2. \ \checkmark$$

Finally, $$\sum_{k=1}^{N} f(k) = \sum_{k=1}^{N}(k+1) = \frac{N(N+1)}{2} + N = \frac{N(N+3)}{2}.$$ For $N = 12$ this is $\tfrac{12 \cdot 15}{2} = 90 < 100$, while $N = 13$ gives $\tfrac{13 \cdot 16}{2} = 104$. So $N = 12$.

Answer 12

Solution: IOQM 2025 Part SEP, Q7

Set $h(x) = f(x) - 4g(x)$. It has degree at most $2$, and the two given ratios say $h(-2) = h(3) = 0$. A polynomial of degree at most $2$ with those two roots is $$h(x) = k(x+2)(x-3)$$ for some constant $k$.

Evaluate at $x = 7$, where all three values are given: $$h(7) = f(7) - 4g(7) = 12 - 4(-6) = 36, \qquad k \cdot 9 \cdot 4 = 36, \qquad k = 1.$$

Now evaluate at $x = 5$: $$f(5) = 4g(5) + h(5) = 4 \cdot 2 + 7 \cdot 2 = 8 + 14 = 22.$$

Neither $f$ nor $g$ was ever determined, and neither needed to be. When two polynomials are tied together only through their ratio, the difference that the ratio kills is usually the object to work with.

Answer 22

Solution: IOQM 2026, Q11

If $P$ has degree $n \ge 1$, then $P(x^2)$ has degree $2n$ and $P(P(x))$ has degree $n \cdot n$. So $n^2 = 2n$, and $n = 2$. Write $P(x) = ax^2 + bx + c$ with $a \ne 0$.

Now subtract, keeping $P(x)$ whole wherever it appears: $$\begin{aligned} P(x^2) - P(P(x)) &= a\bigl(x^4 - P(x)^2\bigr) + b\bigl(x^2 - P(x)\bigr) \\ &= \bigl(x^2 - P(x)\bigr)\bigl(a(x^2 + P(x)) + b\bigr). \end{aligned}$$ This has to be the zero polynomial. A product of two non-zero polynomials is never the zero polynomial, since its leading term is the product of their leading terms, so one factor is zero.

Either $P(x) = x^2$, or $a(x^2 + P(x)) + b = 0$ for every $x$. The second reads $$(a^2 + a)x^2 + abx + (ac + b) = 0,$$ so $a^2 + a = 0$, which with $a \ne 0$ gives $a = -1$; then $ab = 0$ gives $b = 0$, and $ac + b = 0$ gives $c = 0$. That is $P(x) = -x^2$.

Both work: for $P(x) = x^2$ each side is $x^4$, and for $P(x) = -x^2$ each side is $-x^4$. There are $2$ such polynomials.

Answer 2

Solution: PRMO 2012, Q18

Put $n = [x]$, so $x^2 = 7n - 5$. The right-hand side must be non-negative, and a negative $x$ is impossible: it would force $n \le -1$ and then $7n - 5 < 0$. So $x = \sqrt{7n-5}$ with $n \ge 1$, and the condition to check is $n \le \sqrt{7n-5} < n+1$.

Squaring, that condition is $n^2 \le 7n - 5 < (n+1)^2$, which is a test on whole numbers and needs no square roots at all: $$\begin{array}{r|c|c|c|l} n & 7n-5 & n^2 & (n+1)^2 & \\ \hline 1 & 2 & 1 & 4 & \checkmark \\ 2 & 9 & 4 & 9 & \times \\ 3 & 16 & 9 & 16 & \times \\ 4 & 23 & 16 & 25 & \checkmark \\ 5 & 30 & 25 & 36 & \checkmark \\ 6 & 37 & 36 & 49 & \checkmark \\ 7 & 44 & 49 & 64 & \times \end{array}$$ The two failures at $n = 2$ and $n = 3$ are the cases where $7n-5$ reaches $(n+1)^2$ exactly, and beyond $n = 7$ the root falls short of $n$ for good, since $n^2 - 7n + 5 > 0$ from there on.

The four roots are $\sqrt2$, $\sqrt{23}$, $\sqrt{30}$, $\sqrt{37}$, and the sum of their squares is $$2 + 23 + 30 + 37 = 92.$$

Answer 92

Solution: PRMO 2013, Q16

Let $r$ be a common root. From $g(r) = 0$ we get $r^2 + br - 3 = 0$, and $r = 0$ is impossible since it would give $-3 = 0$. So $$b = \frac{3 - r^2}{r}.$$

Substituting into $f(r) = r^3 - 3r + b = 0$ and multiplying by $r$, $$r^4 - 3r^2 + 3 - r^2 = 0, \qquad \text{that is} \qquad r^4 - 4r^2 + 3 = 0,$$ which factors as $\left(r^2 - 1\right)\left(r^2 - 3\right) = 0$. So $r = \pm 1$ or $r = \pm\sqrt3$, and the corresponding values of $b$ are $$r = 1: \ b = 2, \qquad r = -1: \ b = -2, \qquad r = \pm\sqrt3: \ b = 0.$$

Each really occurs: at $b = 2$ the two polynomials share the root $1$, at $b = -2$ they share $-1$, and at $b = 0$ they share $\sqrt3$, since then $f(x) = x(x^2-3)$ and $g(x) = x^2 - 3$.

The distinct values are $2$, $-2$ and $0$, and their sum is $$0.$$

Answer 0

Solution: IOQM 2020, Q11

Suppose some real number $r$ is a zero of both polynomials, so that $$r^2 + ar + b = 0 \qquad \text{and} \qquad r^3 + br + a = 0.$$ Two equations in the unknowns $a$ and $b$ invite elimination, and the first is already solved for $b$, namely $b = -r^2 - ar$. Substituting that into the second and expanding, $$r^3 + (-r^2 - ar)r + a = r^3 - r^3 - ar^2 + a = a(1 - r^2),$$ so the condition for a common root is $a(1 - r^2) = 0$. That one line does most of the work: every solution falls into one of three families: $a = 0$, or $r = 1$, or $r = -1$. We count each in turn and then reconcile the overlaps.

The family $a = 0$

The quadratic becomes $x^2 + b$, which has a real zero exactly when $b \leq 0$, the zeros being $\pm\sqrt{-b}$. The cubic becomes $x^3 + bx = x(x^2 + b)$, which shares those zeros. So every $b$ in $\{-5, -4, -3, -2, -1, 0\}$ succeeds, giving $6$ pairs.

The family $r = 1$

Substituting $r = 1$ into either equation gives the same condition $1 + a + b = 0$, so $b = -1-a$. Requiring $b$ to lie in $X$ restricts $a$ to $\{-5, -4, \ldots, 4\}$, which is $10$ pairs.

The family $r = -1$

Substituting $r = -1$ gives $1 - a + b = 0$ in both, so $b = a - 1$, and requiring $b \in X$ restricts $a$ to $\{-4, -3, \ldots, 5\}$, another $10$ pairs.

These families are not disjoint, and it turns out they meet in a single point. Setting $-1-a = a-1$ forces $a = 0$ and $b = -1$, and that same pair $(0,-1)$ also satisfies the condition of the first family. So $(0,-1)$ lies in all three families and in each of the three pairwise intersections, and inclusion-exclusion gives

$$|S| = 6 + 10 + 10 - 1 - 1 - 1 + 1 = 24.$$

Answer 24

Solution: IOQM 2020, Q21

Write $a$, $b$, $c$ for the three ages in the first year and $S = a+b+c$ for their total, so that in the first year $A$ receives $Na/S$, and so on. Five years later, at the sixth payment, the ages are $a+5$, $b+5$, $c+5$ and their total is $S+15$.

The first clue is that $A$ received half the total in the first year, so $Na/S = N/2$, giving $a = S/2$ immediately. The second clue compares $A$’s sixth payment with her first: $$\frac{N(a+5)}{S+15} = \frac67 \cdot \frac{Na}{S} = \frac67 \cdot \frac{N}{2} = \frac{3N}{7},$$ so $\dfrac{a+5}{S+15} = \dfrac37$. Substituting $a = S/2$ and clearing denominators, $7\left(\tfrac{S}{2}+5\right) = 3(S+15)$, which simplifies to $S/2 = 10$ and hence $S = 20$. So $a = 10$ and $b + c = 10$.

The third clue, that $C$ received twice her first-year amount, isolates $c$: $$\frac{N(c+5)}{35} = 2 \cdot \frac{Nc}{20} = \frac{Nc}{10}, \qquad \text{so} \qquad 10(c+5) = 35c,$$ giving $25c = 50$ and $c = 2$, and therefore $b = 8$. Notice that all three of these clues were ratios, so $N$ has cancelled every time and the ages were determined without it.

Only now do we use the one clue that mentions an actual sum of money. Working in thousands of rupees, $B$ received one thousand less at the sixth payment than at the first: $$\frac{13N}{35} = \frac{8N}{20} - 1 = \frac{14N}{35} - 1.$$ Hence $1 = N/35$ and $N = 35$.

Answer 35

Solution: IOQM 2021 Part A, Q10

Building the polynomial

Write $\alpha = \sqrt7 + \sqrt5$. The first thing to notice is that $$(\sqrt7 + \sqrt5)(\sqrt7 - \sqrt5) = 7 - 5 = 2, \qquad \text{so} \qquad \sqrt7 - \sqrt5 = \frac{2}{\alpha}.$$ The condition $P(\alpha) = 2(\sqrt7 - \sqrt5)$ therefore reads $P(\alpha) = \dfrac{4}{\alpha}$. That is not yet a polynomial condition, but it is close to one: multiplying by $\alpha$ gives $\alpha P(\alpha) = 4$.

Next find what $\alpha$ satisfies. Squaring, $\alpha^2 = 12 + 2\sqrt{35}$, so $\alpha^2 - 12 = 2\sqrt{35}$ and squaring again gives $\alpha^4 - 24\alpha^2 + 144 = 140$, that is $$\alpha^4 - 24\alpha^2 + 4 = 0.$$

Now read that backwards as a statement about the number $4$: $$4 = 24\alpha^2 - \alpha^4 = \alpha\left(24\alpha - \alpha^3\right).$$ Comparing with $\alpha P(\alpha) = 4$ and cancelling $\alpha$, which is certainly not zero, $$P(x) = 24x - x^3.$$ This has integer coefficients and degree three, and $P(2) = 48 - 8 = 40$.

Why degree three is really necessary

The question asks for the polynomial of least degree, so the construction alone does not finish it. Everything below is about ruling out degrees one and two, and the degree-zero case as well.

The whole difficulty is one fact: $\alpha$ satisfies no equation with rational coefficients of degree below four. That comes down to showing that $x^4 - 24x^2 + 4$ does not factor over the rationals. We will justify that implication after checking the factorisations.

It has no rational root. To see why only integer candidates are needed, write a rational root as $h/k$ in lowest terms with $k>0$. Clearing denominators gives $h^4-24h^2k^2+4k^4=0$, so $k$ divides $h^4$. Coprimality forces $k=1$. Then $h$ divides $4$ by the same equation, and $x = 1, 2, 4$ give $-19$, $-76$ and $-124$, with the negative values the same because only even powers appear. So the only possible factorisation is into two quadratics, and being monic with no cubic term they must take the form $$\left(x^2 + ux + v\right)\left(x^2 - ux + w\right) = x^4 + (v + w - u^2)x^2 + u(w-v)x + vw.$$ Matching coefficients gives $u(w-v) = 0$, $v + w - u^2 = -24$ and $vw = 4$, all rational.

• If $u = 0$, then $v + w = -24$ and $vw = 4$, so $v$ and $w$ are roots of $t^2 + 24t + 4$, whose discriminant $576 - 16 = 560$ is not the square of a rational.

• If $w = v$, then $v^2 = 4$ gives $v = \pm 2$, and $2v - u^2 = -24$ makes $u^2$ equal to $28$ or $20$, neither the square of a rational.

So $F(x)=x^4-24x^2+4$ does not factor over the rationals. Suppose a non-zero rational polynomial $Q$ of degree below $4$ also vanished at $\alpha$. Repeated polynomial division, just as in the Euclidean algorithm for integers, reduces degrees and finds a common divisor of $F,Q$. That divisor must be constant: any non-constant divisor of the unfactorable $F$ has degree $4$, too large to divide $Q$. Running the divisions backwards therefore gives rational polynomials $U,V$ with $UF+VQ=1$. Substituting $x=\alpha$ gives $0=1$, a contradiction. Thus no non-zero rational polynomial of degree below $4$ vanishes at $\alpha$.

Now suppose some $P$ of degree at most two worked. Then $xP(x) - 4$ would be a non-zero polynomial of degree at most three vanishing at $\alpha$, which we have just ruled out. So degree three is least, and $P(x) = 24x - x^3$ is the polynomial the question asks for, with $P(2) = 40$.

Answer 40

Solution: IOQM 2022, Q14

Write $s = x + y + z$, so each denominator is $s$ minus the corresponding numerator’s variable, and set $T = \sum \dfrac{1}{s-x}$, the sum running cyclically.

Take the three givens in turn. Since $\dfrac{x}{s-x} = \dfrac{s}{s-x} - 1$, the first says $$sT - 3 = 9, \qquad \text{so} \qquad sT = 12.$$ For the second, note $x^2 = s^2 - (s-x)(s+x)$, so $\dfrac{x^2}{s-x} = \dfrac{s^2}{s-x} - (s+x)$ and summing gives $s^2 T - (3s + s) = 64$, that is $s^2T - 4s = 64$. Using $sT = 12$ this reads $12s - 4s = 64$, so $$s = 8, \qquad T = \tfrac{12}{8} = \tfrac32.$$ For the third, $x^3 = s^3 - (s-x)(s^2+sx+x^2)$ gives $s^3T - \left(3s^2 + s^2 + \sum x^2\right) = 488$. With $s = 8$ and $T = \tfrac32$ this is $768 - 256 - \sum x^2 = 488$, so $\sum x^2 = 24$.

Two of the three elementary symmetric functions follow at once. From $\sum x^2 = s^2 - 2\sum xy$ we get $\sum xy = \tfrac{64-24}{2} = 20$. For the third, evaluate $T$ directly: putting everything over the common denominator $\prod (s-x)$, $$T = \frac{\sum (s-y)(s-z)}{\prod (s-x)} = \frac{s^2 + \sum xy}{s^3 - s^2 \cdot s + s\sum xy - xyz} = \frac{64 + 20}{160 - xyz}.$$ Setting this equal to $\tfrac32$ gives $168 = 3(160 - xyz)$, so $xyz = 104$.

The requested quantity is now immediate, since $\dfrac{x}{yz} + \dfrac{y}{zx} + \dfrac{z}{xy} = \dfrac{x^2+y^2+z^2}{xyz}$: $$\frac{24}{104} = \frac{3}{13}.$$ Hence $m = 3$, $n = 13$ and $m + n = 16$.

Answer 16

Solution: IOQM 2022, Q16

Rearrange each equation to isolate a variable. From $3ab + 2 = 6b$ we get $3b(a-2) = -2$, so $$b = \frac{2}{3(2-a)}.$$ From $3bc + 2 = 5c$ we get $c(3b-5) = -2$, so $c = \dfrac{2}{5-3b}$, and from $3ca + 2 = 4a$ we get $a = \dfrac{2}{4-3c}$.

Now walk round the cycle. Substituting $3b = \dfrac{2}{2-a}$, $$5 - 3b = \frac{5(2-a) - 2}{2-a} = \frac{8-5a}{2-a}, \qquad \text{so} \qquad c = \frac{2(2-a)}{8-5a}.$$ Then $3c = \dfrac{6(2-a)}{8-5a}$ and $$4 - 3c = \frac{4(8-5a) - 6(2-a)}{8-5a} = \frac{20 - 14a}{8-5a},$$ so the last equation becomes $a = \dfrac{2(8-5a)}{20-14a}$. Cross-multiplying, $$a(20 - 14a) = 16 - 10a \quad\Longrightarrow\quad 7a^2 - 15a + 8 = 0,$$ which factors as $(7a - 8)(a - 1) = 0$. The two promised values of the product come from these two roots.

If $a = 1$ then $b = \tfrac{2}{3}$ and $c = \tfrac{2}{3}$, so $abc = \tfrac49$. If $a = \tfrac87$ then $b = \tfrac79$ and $c = \tfrac34$, so $abc = \tfrac87 \cdot \tfrac79 \cdot \tfrac34 = \tfrac23$.

Both are already in lowest terms, so $\tfrac{r}{s} = \tfrac49$ and $\tfrac{t}{u} = \tfrac23$, giving $r + s + t + u = 4 + 9 + 2 + 3 = 18$.

Answer 18

Solution: IOQM 2025 Part SEP, Q7

The three entries are consecutive integers, so the outer two differ by $2$: $$a(n+1)^2 - (a+1)n^2 = 2.$$ Expanding, the $an^2$ terms cancel and this becomes $$n^2 - 2an - a + 2 = 0. \tag{$*$}$$

Now, why must $n$ be an integer? The middle entry $m^2$ is an integer and so is the first, $(a+1)n^2$. From $(*)$ we have $n^2 = 2an + a - 2$, so $$(a+1)n^2 = (a+1)(2an + a - 2),$$ which is an integer only if $2a(a+1)n$ is, so $n$ is rational. Write $n=h/k$ in lowest terms with $k>0$. Clearing denominators in $(*)$ gives $h^2-2ahk+(-a+2)k^2=0$, so $k$ divides $h^2$. Since $h,k$ are coprime, $k=1$, and $n$ is an integer.

With $n$ an integer, solve $(*)$ for $a$: $$a(2n+1) = n^2 + 2, \qquad a = \frac{n^2+2}{2n+1}.$$ Multiply by $4$ to clear the denominator neatly: $4(n^2+2) = (2n+1)(2n-1) + 9$, so $2n+1$ must divide $9$. Hence $2n + 1 \in \{\pm1, \pm3, \pm9\}$ and $n \in \{0, 1, 4, -1, -2, -5\}$.

The negative values of $n$ all make $a$ negative, which is not allowed, so three cases remain: $$\begin{array}{c|c|c} n & a & \text{the three consecutive integers} \\ \hline 0 & 2 & 0,\ 1,\ 2 \\ 1 & 1 & 2,\ 3,\ 4 \\ 4 & 2 & 48,\ 49,\ 50 \end{array}$$ All three are genuine, and the largest middle entry is $$m^2 = 49.$$

Answer 49

Solution: IOQM 2025 Part SEP, Q7

First factor the divisor: $$x^4+x^3+2x^2+x+1 = (x^2+1)(x^2+x+1),$$ which one checks by expanding. Now reduce $x^{2025}$ modulo each factor.

Modulo $x^2+1$ we have $x^2 \equiv -1$, so $$x^{2025} = \left(x^2\right)^{1012} \cdot x \equiv (-1)^{1012}x = x.$$ Modulo $x^2+x+1$ we have $x^3 \equiv 1$, because $x^3 - 1 = (x-1)(x^2+x+1)$, and $2025$ is divisible by $3$, so $$x^{2025} = \left(x^3\right)^{675} \equiv 1.$$

The remainder $R$ has degree at most $3$ and satisfies both congruences. Write $R(x) = x + (x^2+1)(\alpha x + \beta)$, which automatically handles the first. For the second, work modulo $x^2+x+1$, where $x^2 \equiv -x-1$ and therefore $x^2 + 1 \equiv -x$: $$x + (-x)(\alpha x + \beta) = x - \alpha x^2 - \beta x \equiv x + \alpha(x+1) - \beta x = (1 + \alpha - \beta)x + \alpha.$$ This must be the constant $1$, so $\alpha = 1$ and $\beta = 2$. Hence $$R(x) = x + (x^2+1)(x+2) = x^3 + 2x^2 + 2x + 2,$$ and $$R(3) = 27 + 18 + 6 + 2 = 53.$$

Answer 53

Solution: IOQM 2026, Q21

Move the lone unknown in each equation to the right: $$x^2 - xy + y^2 = -z, \qquad y^2 - yz + z^2 = -x, \qquad z^2 - zx + x^2 = -y.$$ Subtracting the third from the first, $y^2 - z^2 - xy + xz = y - z$, which factorises as $$(y - z)(y + z - x - 1) = 0.$$ In the same way the second minus the first gives $(z - x)(z + x - y - 1) = 0$, and the third minus the second gives $(x - y)(x + y - z - 1) = 0$.

All three different. Then $y + z - x$, $z + x - y$ and $x + y - z$ all equal $1$. Adding the first two gives $2z = 2$, so $z = 1$, and in the same way $x = y = 1$, which is not three different numbers. This case is empty.

All three equal, say to $t$. The first equation reads $t^2 = -t$, so $t = 0$ or $t = -1$, and the product $(t^2 - t)^3$ is $0$ or $2^3 = 8$.

Exactly two equal. Renaming $x \to y \to z \to x$ carries each equation to the next and leaves the product unchanged, so it is enough to take $y = z$ with $x$ different. Then $(x - y)(x + y - z - 1) = 0$ gives $x = 1$, and the second equation becomes $1 + 2y^2 = y^2$, that is $y^2 = -1$. The other two equations then hold as well. No real number has square $-1$, which is why the question allows complex numbers, but nothing beyond $y^2 = -1$ is needed: $$(x^2 - y)(y^2 - z)(z^2 - x) = (1 - y)(-1 - y)(-2) = 2(1 - y^2) = 4.$$

Every value of the product is real, so the modulus is the ordinary size, and the distinct values are $0$, $8$ and $4$. Their sum is $12$, with or without the $0$.

Answer 12