Diophantine Equations

Problems

Solutions

Solution: PRMO 2012, Q3

For each positive integer $y$ the value $x = 100 - 3y$ is determined, and it is positive exactly when $3y \le 99$, that is $y \le 33$. Every such $y$ gives a valid pair, so there are $$33$$ pairs, from $(97, 1)$ to $(1, 33)$.

Answer 33

Solution: IOQM 2026, Q10

Zero is a square, $0 = 0^2$, so a sum of two squares may use it. The squares up to $20$ are $0, 1, 4, 9, 16$, and their sums in pairs, repeats allowed, that do not pass $20$ are $$\begin{array}{llll} 0 = 0 + 0, & 1 = 0 + 1, & 2 = 1 + 1, & 4 = 0 + 4,\\ 5 = 1 + 4, & 8 = 4 + 4, & 9 = 0 + 9, & 10 = 1 + 9,\\ 13 = 4 + 9, & 16 = 0 + 16, & 17 = 1 + 16, & 18 = 9 + 9,\\ 20 = 4 + 16. & & & \end{array}$$ The pairs $9 + 16$ and $16 + 16$ pass $20$, so the list is complete, and it holds $13$ integers.

Answer 13

Solution: PRMO 2012, Q4

Write $p = R \times O$. The first equation says $M p = 240$, so $M = 240/p$, and substituting into the second, $$p + \frac{240}{p} = 46, \qquad p^2 - 46p + 240 = 0, \qquad p = 40 \ \text{ or } \ 6.$$

Take the cases in turn. If $p = 40$ then $M = 6$, and the third equation reads $R + 6O = 64$ with $RO = 40$, so $\tfrac{40}{O} + 6O = 64$, that is $3O^2 - 32O + 20 = 0$, whose roots are $10$ and $\tfrac23$. Only $O = 10$ is a whole number, giving $R = 4$.

If $p = 6$ then $M = 40$ and $R + 40O = 64$ with $RO = 6$, leading to $20O^2 - 32O + 3 = 0$, whose roots $\tfrac32$ and $\tfrac1{10}$ are both fractions.

So $(R, M, O) = (4, 6, 10)$, which satisfies all three equations, and $$R + M + O = 20.$$

Answer 20

Solution: PRMO 2015 Part A, Q2

Multiplying $a + b = \dfrac{a}{b} + \dfrac{b}{a}$ through by $ab$ gives $$ab(a+b) = a^2 + b^2.$$ Now write $d = \gcd(a,b)$ and $a = dx$, $b = dy$ with $\gcd(x,y) = 1$. Substituting and cancelling $d^2$ from both sides, $$d\,xy(x+y) = x^2 + y^2.$$ This is where coprimality does the work. The number $x$ divides the left-hand side, so it divides $x^2 + y^2$, and since it certainly divides $x^2$ it must divide $y^2$. But $x$ and $y$ share no factor, so $x = 1$. The same argument on $y$ gives $y = 1$.

With $x = y = 1$ the equation reads $2d = 2$, so $d = 1$ and therefore $a = b = 1$. A quick check confirms it: $1 + 1 = \tfrac11 + \tfrac11$. Hence $$a^2 + b^2 = 2.$$

Answer 2

Solution: PRMO 2017, Q8

Let $p$ be the number of pens and $q$ the number of notebooks, so $11p + 13q = 1000$ with $p, q \ge 0$.

Work modulo $13$. Since $1000 = 76 \times 13 + 12$, the equation reads $11p \equiv 12$, and $11 \equiv -2$, so $-2p \equiv 12$, that is $p \equiv -6 \equiv 7 \pmod{13}$. Writing $p = 7 + 13t$, $$q = \frac{1000 - 11(7 + 13t)}{13} = \frac{923 - 143t}{13} = 71 - 11t.$$

Both are non-negative exactly when $0 \le t \le \tfrac{71}{11} = 6.45\ldots$, that is $t = 0, 1, \ldots, 6$. That is $7$ ways, running from $(p,q) = (7,71)$ to $(85,5)$, and in every one of them the person buys at least one of each.

Answer 7

Solution: PRMO 2017, Q12

Let the numbers of boys and girls be $4k$ and $3k$. On the day in question the numbers present are $4k - 8$ boys and $3k - 14$ girls, and we are told $$4k - 8 = (3k - 14)^2.$$ Expanding, $9k^2 - 84k + 196 = 4k - 8$, that is $$9k^2 - 88k + 204 = 0.$$ The discriminant is $88^2 - 4 \cdot 9 \cdot 204 = 7744 - 7344 = 400$, so $$k = \frac{88 \pm 20}{18}, \qquad k = 6 \ \text{ or } \ k = \frac{68}{18}.$$ Only $k = 6$ is a whole number, so the class has $24$ boys and $18$ girls, a total of $42$.

The answer checks out: $24 - 8 = 16$ boys and $18 - 14 = 4$ girls were present, and $16 = 4^2$.

Answer 42

Solution: PRMO 2018, Q6

From the first equation, $c = a + b - 1$. Substituting into the second, $$a^2 + b^2 - (a+b-1)^2 = -1.$$ Expanding $(a+b-1)^2 = a^2 + b^2 + 1 + 2ab - 2a - 2b$ and cancelling, $$-2ab + 2a + 2b - 1 = -1, \qquad \text{that is} \qquad a + b = ab.$$ Rearranged, $ab - a - b + 1 = 1$, so $$(a-1)(b-1) = 1.$$

Over the integers the only factorisations of $1$ are $1 \times 1$ and $(-1) \times (-1)$, giving $(a,b) = (2,2)$ or $(a,b) = (0,0)$. The corresponding $c$ are $3$ and $-1$, and both really do satisfy the original pair: $4 + 4 - 9 = -1$ and $0 + 0 - 1 = -1$.

The two values of $a^2 + b^2 + c^2$ are $4 + 4 + 9 = 17$ and $0 + 0 + 1 = 1$, and their sum is $$17 + 1 = 18.$$

Answer 18

Solution: IOQM 2025 Part SEP, Q7

The two ages differ by $2025 - 2012 = 13$, so we need squares with $$p^2 - q^2 = 13, \qquad (p-q)(p+q) = 13.$$ As $13$ is prime and $p + q$ is the larger factor, $p - q = 1$ and $p + q = 13$, giving $p = 7$ and $q = 6$. So the person was $36$ in $2012$ and is $49$ in $2025$.

The next perfect cube at or beyond $49$ is $64 = 4^3$, reached $64 - 49 = 15$ years after $2025$. So $m = 15$.

Answer 15

Solution: IOQM 2025 Part SEP, Q7

For a closed cylinder of radius $r$ and height $h$, $$2\pi r^2 + 2\pi r h = \pi r^2 h.$$ Dividing by $\pi r$, which is positive, gives $2r + 2h = rh$, so $$rh - 2r - 2h + 4 = 4, \qquad (r-2)(h-2) = 4.$$

Both $r$ and $h$ are positive integers, and neither $r-2$ nor $h-2$ can be negative here, since a negative factor would force the other to be negative too and then $r, h \le 1$, which fails the equation. So the factorisations of $4$ give $$(r,h) = (3,6), \quad (4,4), \quad (6,3),$$ with volumes $\pi r^2 h$ equal to $54\pi$, $64\pi$ and $108\pi$.

The smallest value of $k$ is $54$.

Answer 54

Solution: IOQM 2026, Q6

Write $n = 10a + b$, where $a$ runs from $1$ to $9$ and $b$ from $0$ to $9$. The condition is $10a + b = 26 + ab$. Gathering everything that involves $a$ on one side, $$10a - ab = 26 - b, \qquad a(10 - b) = 26 - b.$$ The right-hand side is $(10 - b) + 16$, so subtracting $10 - b$ from both sides gives $$(a - 1)(10 - b) = 16.$$

Now $a - 1$ lies between $0$ and $8$, and $10 - b$ between $1$ and $10$. Of the ways to write $16$ as a product of two positive integers, $1 \times 16$ and $16 \times 1$ each have a factor out of range, which leaves three: $$2 \times 8,\quad 4 \times 4,\quad 8 \times 2,$$ giving $(a, b) = (3, 2), (5, 6), (9, 8)$. Each checks: $32 = 26 + 6$, $56 = 26 + 30$ and $98 = 26 + 72$. There are $3$ such numbers.

Answer 3

Solution: PRMO 2012, Q5

Since $n^2 + 20n + 12 = (n+10)^2 - 88$, asking for $S_n = k^2$ is asking for $$(n+10)^2 - k^2 = 88, \qquad (n + 10 - k)(n + 10 + k) = 88.$$

The two factors have the same parity, since they differ by $2k$, and their product $88$ is even, so both are even. Writing $88 = 2 \times 44 = 4 \times 22$ (the split $8 \times 11$ mixes parities), $$n + 10 = \frac{2 + 44}{2} = 23 \implies n = 13,$$ $$n + 10 = \frac{4 + 22}{2} = 13 \implies n = 3.$$ Both work: $S_{13} = 441 = 21^2$ and $S_3 = 81 = 9^2$.

The sum of all possible $n$ is $$13 + 3 = 16.$$

Answer 16

Solution: PRMO 2013, Q7

Let Arthur have $a$ marbles and Beatrice $b$, with $a + b = n$.

Arthur’s statement is that for some whole number $x \ge 1$, giving away $x$ leaves $b + x = 2(a - x)$, that is $$3x = 2a - b.$$ So $2a - b$ must be a positive multiple of $3$, at least $3$.

Beatrice’s is that for some $y \ge 1$, $a + y = 3(b - y)$, that is $$4y = 3b - a,$$ so $3b - a$ is a positive multiple of $4$, at least $4$.

Now write both in terms of $n$. Since $b = n - a$, $$2a - b = 3a - n, \qquad 3b - a = 4b - n.$$ The first is a multiple of $3$ exactly when $3 \mid n$, and the second is a multiple of $4$ exactly when $4 \mid n$. So $12 \mid n$.

And $n = 12$ is achievable: take $a = 5$ and $b = 7$. Then $2a - b = 3$, so Arthur gives one marble and Beatrice is left with $8$ against Arthur’s $4$; and $3b - a = 16$, so Beatrice gives four marbles and Arthur has $9$ against Beatrice’s $3$. Both statements hold, so $$n = 12.$$

Answer 12

Solution: PRMO 2013, Q10

The statement never says what kind of numbers Carol was given, and that matters: over the positive reals the answer is not unique. Taking $a = \tfrac12$, $b = 1$ and $c = 1$ satisfies the condition, since $1 + \tfrac12 = \tfrac32 = 1 \cdot \tfrac32$, and the sum is $\tfrac52$ rather than $3$. The intended reading, and the one that gives a single answer, is that the three numbers are positive integers, and that is what is assumed below.

Let the three numbers be $a \le b \le c$, positive integers, with $c$ the largest. Carol should have computed $c + ab$ and instead computed $c(a+b)$, and the two agreed: $$c + ab = c(a+b).$$ Solving for $c$, $$c\bigl(1 - a - b\bigr) = -ab, \qquad c = \frac{ab}{a + b - 1}.$$

Now use the fact that $c$ is the largest, so $c \ge b$: $$\frac{ab}{a+b-1} \ge b \implies a \ge a + b - 1 \implies b \le 1.$$ So $b = 1$, and since $a \le b$ we may as well take $a \le 1$ too, giving $a = 1$. Then $$c = \frac{1}{1 + 1 - 1} = 1.$$

The three numbers are $1$, $1$, $1$, and indeed $1 + 1 \times 1 = 2 = 1 \times (1+1)$. Their sum is $$3.$$

Answer 3

Solution: PRMO 2013, Q14

Write $T_k = 1 + 2 + \cdots + k = \tfrac12 k(k+1)$. Nothing needs to be known about which triangular numbers are squares in general, because both answers are found by looking at the first few.

The smallest odd index is immediate: $T_1 = 1 = 1^2$, so $m = 1$.

For the smallest even index, take the even values in turn: $$T_2 = 3, \qquad T_4 = 10, \qquad T_6 = 21, \qquad T_8 = 36 = 6^2.$$ None of the first three is a square, and the fourth is, so $n = 8$. Hence $$m + n = 1 + 8 = 9.$$

A word on $m = 1$, which surprises some readers. The sum $1 + 2 + \cdots + m$ with $m = 1$ is just $1$, and $1$ is a square, so the condition is met. If one insists on at least two terms, the smallest odd index becomes $49$ and the answer would be $57$; the question as printed does not say that, so the answer is $9$.

Answer 9

Solution: PRMO 2014, Q14

Let there be $n$ people, so the family drank $8n$ ounces in all. Write $M$ for the total milk and $C$ for the total coffee, so $M + C = 8n$.

Megan drank eight ounces, made of $\tfrac17$ of the milk and $\tfrac{2}{17}$ of the coffee: $$\frac{M}{7} + \frac{2C}{17} = 8, \qquad \text{that is} \qquad 17M + 14C = 952.$$

Substituting $M = 8n - C$ gives $136n - 3C = 952$, so $$C = \frac{136n - 952}{3}, \qquad M = 8n - C = \frac{952 - 112n}{3}.$$

Both totals are positive, since every cup contained some of each: $$C > 0 \implies n > 7, \qquad M > 0 \implies n < 8.5.$$ The only integer between is $$n = 8.$$ Then $C = \tfrac{136}{3}$ and $M = \tfrac{56}{3}$ ounces, which do sum to $64 = 8 \times 8$.

That $n = 8$ is forced does not by itself show that eight people can actually be served, so here is a distribution that works. Megan takes $\tfrac17$ of the milk and $\tfrac{2}{17}$ of the coffee, namely $\tfrac83$ ounces of milk and $\tfrac{16}{3}$ of coffee, which is eight ounces. That leaves $16$ ounces of milk and $40$ of coffee for the other seven, so give each of them $\tfrac{16}{7}$ of milk and $\tfrac{40}{7}$ of coffee, again eight ounces. Every cup contains some of each, as the question requires.

The fractions $\tfrac17$ and $\tfrac{2}{17}$ were chosen to squeeze the answer between two consecutive integers, which is the whole mechanism: one drinker’s share of each ingredient bounds the number of drinkers from both sides.

Answer 8

Solution: PRMO 2015 Part A, Q10

There are two cases, and one of them is quickly dismissed. If $12$ were the hypotenuse then both legs are less than $12$ and the perimeter is under $36$, which will turn out to be far from the best. So take $12$ as a leg, with the other leg $b$ and hypotenuse $c$, both positive integers. Then $b^2 + 144 = c^2$, that is $$(c-b)(c+b) = 144.$$ The perimeter is $12 + b + c$, so we want $b + c$ as large as possible. Writing $c - b = u$ and $c + b = v$ with $uv = 144$, we want $v$ as large as possible, hence $u$ as small as possible.

Two constraints bite. First, $u$ and $v$ have the same parity, because $u + v = 2c$ is even, and since their product $144$ is even they must both be even. Second, $v > u$, so that $b > 0$. The smallest even value of $u$ is $2$, giving $v = 72$, and then $$c = \frac{u+v}{2} = 37, \qquad b = \frac{v-u}{2} = 35.$$ A check confirms the triangle: $12^2 + 35^2 = 144 + 1225 = 1369 = 37^2$. The perimeter is $12 + 35 + 37 = 84$.

Why does $u = 1$ fail, one might reasonably ask, since it would give the larger $v = 144$. It fails because then $c = \tfrac{145}{2}$ is not an integer. The parity condition is doing real work, and it is the step most often skipped.

Answer 84

Solution: PRMO 2019, Q9

For any fraction $p/q$ with $q < 100$, $$\left|\frac{p}{q} - \frac{22}{7}\right| = \frac{|7p - 22q|}{7q}.$$ The numerator $|7p - 22q|$ is a non-negative integer, and it is zero only when $p/q$ equals $22/7$, which is excluded. So it is at least $1$, and $$\left|\frac{p}{q} - \frac{22}{7}\right| \ge \frac{1}{7q} \ge \frac{1}{7 \times 99} = \frac{1}{693}.$$

Equality needs both $q = 99$ and $|7p - 22q| = 1$, so $7p = 22 \times 99 \pm 1 = 2178 \pm 1$. Of the two, $2177 = 7 \times 311$ is divisible by $7$ while $2179$ is not. Hence the unique closest fraction is $$\frac{p}{q} = \frac{311}{99},$$ and it does sit at distance exactly $1/693$, so the bound is attained. Finally $$p - 3q = 311 - 297 = 14.$$

The reason the search space collapses so fast is worth noting. Denominators near the limit are the only ones that can be good, because a small denominator forces the fraction onto a coarse grid. That is the same principle behind continued fractions, of which $22/7$ is the most famous example.

Answer 14

Solution: PRMO 2019, Q16

First solve $13x + 17y = 10000$ in non-negative integers. Working modulo $17$, and using $10000 = 17 \times 588 + 4$, $$13x \equiv 4 \pmod{17}.$$ Since $13 \equiv -4$, this reads $-4x \equiv 4$, that is $x \equiv -1 \equiv 16 \pmod{17}$. So $$x = 16 + 17t, \qquad y = \frac{10000 - 13(16+17t)}{17} = 576 - 13t,$$ and both are non-negative exactly for $0 \le t \le 44$.

Now measure the gap: $$x - y = (16 + 17t) - (576 - 13t) = 30t - 560.$$ This vanishes at $t = 18\tfrac23$, so the integer candidates are $t = 18$, giving $x - y = -20$, and $t = 19$, giving $x - y = 10$. The minimum of $|x-y|$ is therefore $10$, attained at $t = 19$: $$x = 339, \qquad y = 329, \qquad 13 \times 339 + 17 \times 329 = 10000.$$

The following year the school buys $y$ pens and $x$ notebooks, so it pays $13y + 17x$. The difference is $$(13y + 17x) - (13x + 17y) = 4(x - y) = 4 \times 10 = 40.$$ The school paid £$40$ more, and one may check it directly: $13 \times 329 + 17 \times 339 = 4277 + 5763 = 10040$.

The identity $4(x-y)$ is the whole reason the question insisted that $x$ and $y$ be as close as possible. Swapping the two quantities in a purchase always costs the price difference times the quantity difference, here $(17 - 13)(x - y)$.

Answer 40

Solution: IOQM 2022, Q4

The successive values show how the starting numbers and added constants grow: $$\begin{array}{c|c|c} \text{move}&\text{Alice}&\text{Bob}\\\hline 1&3M+2&2N+27\\ 2&9M+8&4N+81\\ 3&27M+26&8N+189\\ 4&81M+80&16N+405 \end{array}$$ Setting them equal, $$81M + 80 = 16N + 405, \qquad \text{that is} \qquad 81M - 16N = 325.$$

We want the smallest $M + N$ in positive integers, so read the equation modulo $16$, where the awkward coefficient becomes harmless: $81 \equiv 1$, and $325 = 20 \cdot 16 + 5$, so $$M \equiv 5 \pmod{16}.$$ So $M = 5 + 16t$ for some integer $t \ge 0$, and substituting back gives $16N = 81(5+16t) - 325 = 80 + 1296t$, that is $N = 5 + 81t$. Hence $$M + N = 10 + 97t,$$ which is smallest at $t = 0$, where $M = N = 5$ and both are positive. The minimum is $M + N = 10$.

Answer 10

Solution: IOQM 2022, Q5

Eleven consecutive squares suggest centring the run, so write the terms as $(m+5-5)^2, \ldots, (m+5+5)^2$ and let $u = m+5$. The cross terms cancel in pairs and $$\sum_{k=-5}^{5}(u+k)^2 = 11u^2 + 2u\sum_{k=-5}^{5}k + \sum_{k=-5}^{5}k^2 = 11u^2 + 110.$$

So we need $11(u^2 + 10) = n^2$. Since $11$ is prime and divides $n^2$, it divides $n$; writing $n = 11t$ and cancelling gives $$u^2 + 10 = 11t^2, \qquad \text{that is} \qquad u^2 - 11t^2 = -10.$$

Now hunt for the smallest solution with $u > 5$, since $m = u - 5$ must be positive. Testing $t$ in turn, $11t^2 - 10$ must be a perfect square: $t=1$ gives $1$, so $u=1$ and $m = -4$, which is not positive; $t = 2, 3, 4, 5, 6$ give $34, 89, 166, 265, 386$, none of them squares; and $t = 7$ gives $11 \cdot 49 - 10 = 529 = 23^2$.

The first admissible $t$ is also the best one, because the quantity we are minimising is $m + n = (u - 5) + 11t$, and $u = \sqrt{11t^2 - 10}$ increases with $t$, so both terms do.

So $u = 23$, hence $m = 18$, and $n = 11t = 77$. Checking, the eleven squares from $18^2$ to $28^2$ do sum to $5929 = 77^2$. Therefore $m + n = 18 + 77 = 95$.

Answer 95

Solution: IOQM 2022, Q6

Since $a^3 - b^3 = 25 + ab > 0$ we know $a > b$, so the gap $a - b$ is a positive integer, and the useful question is how large it can be.

For fixed $b$, increasing $a$ by $1$ changes $a^3-b^3-ab$ by $3a^2+3a+1-b$, positive when $a>b$. Thus if $a\ge b+2$, its smallest possible value occurs at $a=b+2$, and $$a^3 - b^3 - ab \geq (b+2)^3 - b^3 - b(b+2) = 5b^2 + 10b + 8,$$ which already exceeds $25$ for every $b \geq 2$. That leaves only $b = 1$ with $a \geq 3$, where the condition reads $a^3 - 1 - a = 25$, so $a^3 - a = 26$; but $3^3 - 3 = 24$ and $4^3 - 4 = 60$, so there is no solution. Hence $a = b + 1$.

Substituting $a = b+1$ turns the cubic into a quadratic: $$(b+1)^3 - b^3 - b(b+1) = 3b^2 + 3b + 1 - b^2 - b = 2b^2 + 2b + 1 = 25,$$ so $b^2 + b - 12 = 0$ and $(b+4)(b-3) = 0$. Only $b = 3$ is positive, giving $a = 4$, and indeed $64 - 27 - 12 = 25$.

The solution is unique, so the largest value of $a^2 + b^3$ is the only one: $16 + 27 = 43$.

Answer 43

Solution: IOQM 2022, Q19

Placing plus signs in a string of $1$s produces a sum of repunits, and since $1111 > 1000$ only $1$, $11$ and $111$ can appear. Suppose we use $\alpha$ ones, $\beta$ elevens and $\gamma$ one hundred and elevens, so that $$\alpha + 11\beta + 111\gamma = 1000, \qquad n = \alpha + 2\beta + 3\gamma.$$

Eliminate $\alpha$ using the first equation. Substituting $\alpha = 1000 - 11\beta - 111\gamma$, $$n = 1000 - 9\beta - 108\gamma = 1000 - 9\left(\beta + 12\gamma\right).$$ So $n$ is determined by the single quantity $w = \beta + 12\gamma$, and counting the achievable $n$ is the same as counting the achievable $w$.

The constraint is $\alpha \geq 0$, that is $11\beta + 111\gamma \leq 1000$. For each $\gamma$ from $0$ to $9$ this allows $\beta$ from $0$ up to $\left\lfloor \tfrac{1000 - 111\gamma}{11} \right\rfloor$, and the resulting $w$ fills the interval from $12\gamma$ to $12\gamma + \left\lfloor \tfrac{1000-111\gamma}{11} \right\rfloor$. Their endpoints are all visible in one table: $$\begin{array}{c|rrrrrrrrrr} \gamma&0&1&2&3&4&5&6&7&8&9\\\hline \min w&0&12&24&36&48&60&72&84&96&108\\ \max w&90&92&94&96&98&100&102&104&106&108 \end{array}$$ Each interval through $\gamma=8$ starts before the preceding one ends, so their union contains every integer from $0$ to $106$. The final case $\gamma = 9$ forces $\beta = 0$ and contributes the isolated value $w = 108$.

So $w$ takes $107 + 1 = 108$ values, and $107$ is the one gap. Hence $a = 108$, and the sum of its digits is $1 + 0 + 8 = 9$.

Answer 9

Solution: IOQM 2023, Q3

Since $\alpha$ and $\beta$ are positive integers, the inequality $\dfrac{16}{37} < \dfrac{\alpha}{\beta}$ says $37\alpha - 16\beta > 0$, and an integer greater than zero is at least $1$. The same applies on the other side. So there are positive integers $u$ and $v$ with $$37\alpha - 16\beta = u \ge 1, \qquad 7\beta - 16\alpha = v \ge 1.$$

Now eliminate $\alpha$. Multiplying the first by $16$ and the second by $37$ and adding, $$16u + 37v = (592\alpha - 256\beta) + (259\beta - 592\alpha) = 3\beta.$$ That single line is the whole problem: $3\beta$ must be expressible as $16u + 37v$ with $u$ and $v$ positive integers.

Two constraints then pin $\beta$ down. First, reading the equation modulo $3$ and using $16 \equiv 1$ and $37 \equiv 1$, we get $u + v \equiv 0 \pmod 3$. Second, we want $16u + 37v$ as small as possible. Take $v = 1$: then $u + 1$ must be a multiple of $3$, so $u \ge 2$, and the smallest value is $16 \cdot 2 + 37 = 69$, giving $\beta = 23$. Take $v \ge 2$ instead: then $16u + 37v \ge 16 + 74 = 90 > 69$. So no smaller $\beta$ is possible.

It remains to confirm that $\beta = 23$ really occurs. From $u = 2$ and $v = 1$ we get $37\alpha = 16 \cdot 23 + 2 = 370$, so $\alpha = 10$, and indeed $$\frac{16}{37} < \frac{10}{23} < \frac{7}{16},$$ both by cross-multiplication: $16 \cdot 23 = 368 < 370 = 10 \cdot 37$, and $10 \cdot 16 = 160 < 161 = 7 \cdot 23$.

Answer 23

Solution: IOQM 2023, Q4

Move the $-1$ across to get $$x^4 + 1 = (x-1)(y^3 - 23),$$ The case $x=1$ would give $1=-1$ in the original equation, so $x>1$. Hence $x-1$ divides $x^4+1$. This is the kind of divisibility that collapses immediately, because modulo $x-1$ we have $x \equiv 1$, hence $$x^4 + 1 \equiv 1^4 + 1 = 2 \pmod{x-1}.$$ Therefore $x - 1$ divides $2$, leaving only $x = 2$ and $x = 3$.

Each candidate is now a single line of arithmetic. If $x = 2$ then $17 = y^3 - 23$, so $y^3 = 40$, and $40$ is not a cube. If $x = 3$ then $82 = 2(y^3 - 23)$, so $y^3 = 64$ and $y = 4$, which is a genuine solution.

Only one pair survives, so the maximum of $x + y$ is $3 + 4 = 7$.

Answer 7

Solution: IOQM 2024, Q13

Move everything to one side of $ac + b + c = bc + a + 66$: $$ac - bc - a + b + c = 66.$$ The first four terms group as $c(a-b) - (a-b) = (a-b)(c-1)$, so $$(a-b)(c-1) + c = 66, \qquad \text{that is} \qquad (a-b)(c-1) + (c-1) = 65,$$ and therefore $$(c-1)\,(a - b + 1) = 65.$$ The step of writing $c$ as $(c-1) + 1$ is the one that makes the whole thing factor, and it is worth looking for whenever a stray linear term spoils an otherwise clean grouping.

Since $65 = 5 \cdot 13$, the factor $c - 1$ is one of $1, 5, 13, 65$. Combine each case with $a+b+c = 32$: $$\begin{array}{c|c|c|c} c-1 & c & a - b & (a, b) \\ \hline 1 & 2 & 64 & a+b = 30 \Rightarrow b = -17, \text{ rejected} \\ 5 & 6 & 12 & a+b = 26 \Rightarrow (19, 7) \\ 13 & 14 & 4 & a+b = 18 \Rightarrow (11, 7) \\ 65 & 66 & 0 & c > 32, \text{ rejected} \end{array}$$ Two candidates survive the positivity requirement, and the condition $a > c$ decides between them: $(a,b,c) = (11,7,14)$ has $a < c$, while $(19,7,6)$ has $a > c$ as required. A check confirms it: $19 \cdot 6 + 7 + 6 = 127$ and $7 \cdot 6 + 19 + 66 = 127$.

So $a = 19$.

Answer 19

Solution: IOQM 2025 Part SEP, Q7

For any fraction $\tfrac ab \ne \tfrac34$, $$\left| \frac34 - \frac ab \right| = \frac{|3b - 4a|}{4b}.$$ The numerator is a positive integer, so the distance is at least $\dfrac{1}{4b} \ge \dfrac{1}{60}$ because $b \le 15$.

Can $\dfrac1{60}$ be attained? That needs $b = 15$ and $|3 \cdot 15 - 4a| = 1$, that is $4a = 44$ or $46$. The first gives $a = 11$, and $\gcd(11,15) = 1$, so $$\frac{11}{15} \qquad \text{is at distance} \qquad \left|\frac{45 - 44}{60}\right| = \frac1{60}.$$

Nothing else comes closer, since any other fraction has $b \le 15$ and numerator at least $1$, and equality in $\tfrac1{4b} = \tfrac1{60}$ forces $b = 15$. So $a + b = 11 + 15 = 26$.

Answer 26

Solution: PRMO 2012, Q19

Collect the terms in $x$: $$x^2 - (2y+2)x + \left(4y^2 - 4y - 8\right) = 0.$$ Its discriminant is $$(2y+2)^2 - 4\left(4y^2 - 4y - 8\right) = -12y^2 + 24y + 36,$$ which factorises as $-12(y-3)(y+1)$.

For a real solution this must be non-negative, so $-1 \le y \le 3$, and for an integer solution it must also be a perfect square. Checking the five values:

• $y = -1$: discriminant $0$, giving $x = 0$.

• $y = 0$: discriminant $36$, giving $x = 4$ or $x = -2$.

• $y = 1$: discriminant $48$, not a square.

• $y = 2$: discriminant $36$, giving $x = 6$ or $x = 0$.

• $y = 3$: discriminant $0$, giving $x = 4$.

That is six pairs in all: $(0,-1)$, $(4,0)$, $(-2,0)$, $(6,2)$, $(0,2)$ and $(4,3)$.

Answer 6

Solution: PRMO 2017, Q23

The left-hand side factorises: $$7x^2 - 44x + 12 = (7x - 2)(x - 6).$$ So we need $(7x-2)(x-6) = p^n$. If the convention in force allows $n = 0$, that case dies at once: the product would be $1$, forcing both factors to be $1$ or both to be $-1$, and $7x - 2$ is neither for any integer $x$. So $n \ge 1$ and the right-hand side is a genuine power of a prime.

Each factor is therefore $\pm$ a power of $p$, and the two signs agree since the product is positive. Suppose both factors are divisible by $p$. Then $p$ divides $$(7x - 2) - 7(x - 6) = 40,$$ so $p$ is $2$ or $5$. Those are small, so a large $p$ requires one factor to have no $p$ in it at all, that is to be $\pm1$.

Check the possibilities. The factor $7x - 2$ is $\equiv 5 \pmod 7$, so it is never $\pm1$. That leaves $x - 6 = \pm1$:

• $x = 7$: then $x - 6 = 1$ and $7x - 2 = 47$, so $p^n = 47$, a prime.

• $x = 5$: then $x - 6 = -1$ and $7x - 2 = 33$, giving $-33$, which is negative and so not a prime power.

Hence the largest prime is $$p = 47,$$ attained at $x = 7$ with $n = 1$.

Answer 47

Solution: PRMO 2018, Q15

Write $2a - b = p^2$, $a - 2b = q^2$ and $a + b = r^2$. The first minus the second is the third: $$p^2 - q^2 = r^2, \qquad \text{that is} \qquad q^2 + r^2 = p^2,$$ so $(q, r, p)$ is a Pythagorean triple. Solving the last two equations for $a$ and $b$, $$b = \frac{r^2 - q^2}{3}, \qquad a = \frac{2r^2 + q^2}{3},$$ so we need $r > q$ and $3 \mid r^2 - q^2$, and we are minimising $b$.

Now the arithmetic that decides everything. Squares are $0$ or $1$ modulo $3$, so if neither leg of the triple were a multiple of $3$ we would have $q^2 + r^2 \equiv 2 \pmod 3$, which is not a square. So at least one leg is divisible by $3$. If exactly one is, then $r^2 - q^2 \equiv \pm1 \pmod 3$ and $b$ is not an integer. Hence $$3 \mid q \quad\text{and}\quad 3 \mid r.$$

Write $q = 3Q$, $r = 3R$; then $p = 3P$ and $(Q, R, P)$ is itself a Pythagorean triple, and $$b = \frac{9R^2 - 9Q^2}{3} = 3\left(R^2 - Q^2\right) = 3(R-Q)(R+Q).$$ The smaller leg of a Pythagorean triple is at least $3$: from $Q^2 = P^2 - R^2$ with $P \ge R + 1$ we get $Q^2 \ge 2R + 1 \ge 2Q + 3$, that is $(Q-3)(Q+1) \ge 0$. (It is not $0$, since $q = 0$ would make $p = r$ and the squares would not be distinct.) So $Q \ge 3$, $R \ge Q + 1 \ge 4$ and $R + Q \ge 7$, and $R > Q$ gives $R - Q \ge 1$. Therefore $b \ge 21$, with equality only for $(Q,R) = (3,4)$.

That case does occur: $q = 9$, $r = 12$, $p = 15$ give $$b = 21, \qquad a = \frac{2 \cdot 144 + 81}{3} = 123,$$ and then $2a - b = 225 = 15^2$, $a - 2b = 81 = 9^2$, $a + b = 144 = 12^2$, three distinct squares. So the smallest possible $b$ is $21$.

Answer 21

Solution: IOQM 2020, Q13

Two values work, and finding them costs nothing: at $n = 2$ we get $|4 + 25 - 65| = 36 = 6^2$, and at $n = 4$ we get $|16 + 625 - 65| = 576 = 24^2$. The real content of the problem is showing that the list stops there, so that the answer is $2 + 4 = 6$. We rule out the odd exponents and the even ones by quite different arguments.

No odd exponent works

The case $n = 1$ gives $58$, which is not a square, so take $n$ odd with $n \geq 3$, where the expression is comfortably positive. Working modulo $3$, where both $2$ and $5$ are congruent to $-1$ and $65$ is congruent to $2$, $$2^n + 5^n - 65 \equiv (-1)^n + (-1)^n - 2 \equiv -1 - 1 - 2 \equiv 2 \pmod 3$$ for odd $n$. Every perfect square is congruent to $0$ or $1$ modulo $3$, never to $2$, so no odd exponent can succeed.

No even exponent beyond four works

Let $n$ be even with $n \geq 8$, so that $2^n$ comfortably exceeds $65$. The point is that $5^n$ is itself a perfect square when $n$ is even, namely $\bigl(5^{n/2}\bigr)^2$, and our expression sits just above it: $$2^n + 5^n - 65 = \bigl(5^{n/2}\bigr)^2 + \bigl(2^n - 65\bigr).$$ For the total to be a square it would have to reach at least the next square up, which is $\bigl(5^{n/2} + 1\bigr)^2 = \bigl(5^{n/2}\bigr)^2 + 2 \cdot 5^{n/2} + 1$, and that would require $$2^n - 65 \geq 2 \cdot 5^{n/2} + 1.$$ This is hopeless, because $5^{n/2} = (\sqrt5)^n$ and $\sqrt5 > 2$, so $5^{n/2}$ already exceeds $2^n$ on its own. The gap between consecutive squares up there is wider than the amount $2^n - 65$ we have to spend.

Only $n = 6$ escapes both arguments, and it fails on inspection, since $64 + 15625 - 65 = 15624$, one short of $125^2$. The solutions are $n = 2$ and $n = 4$, and their sum is $6$.

Answer 6

Solution: IOQM 2020, Q29

Write $g = \gcd(a,b)$ and $a = gp$, $b = gq$ with $p > q \geq 1$ and $\gcd(p,q) = 1$. Then $$c = \frac{ab}{a-b} = \frac{g^2pq}{g(p-q)} = \frac{g\,pq}{p-q}.$$ Because $p$ and $q$ are coprime, any common factor of $p-q$ with $p$ would also divide $q$, and similarly the other way round, so $p-q$ is coprime to $pq$. For $c$ to be a whole number, then, $p-q$ must divide $g$ outright. Write $g = (p-q)t$.

Everything is now expressed in three free parameters: $$a = (p-q)tp, \qquad b = (p-q)tq, \qquad c = t\,pq,$$ and the quantity we care about factors pleasantly, $$a+b+c = t\left[(p-q)(p+q) + pq\right] = t\left(p^2 + pq - q^2\right).$$

So we are choosing a coprime pair $p > q$, computing $v = p^2 + pq - q^2$, and then taking $t$ as large as the ceiling permits. No comparison of the possibilities is needed, because the question asks for the largest value of $a+b+c$ not exceeding $99$, and the ceiling itself is reached: the pair $(p,q) = (3,1)$ gives $v = 11$, and $t = 9$ makes $tv = 99$ exactly.

Taking $p = 3$, $q = 1$ and $t = 9$ gives $a = 54$, $b = 18$ and $c = 27$, and indeed $$\frac{54 \cdot 18}{54 - 18} = \frac{972}{36} = 27, \qquad 54 + 18 + 27 = 99.$$

Answer 99

Solution: IOQM 2023, Q11

The right-hand side is a quadratic in $n$ with an awkward middle term, so complete the square after clearing the fraction that would otherwise appear. Multiplying $m^2 = 4n^2 - 5n + 16$ by $16$, $$16m^2 = 64n^2 - 80n + 256 = (8n-5)^2 + 231.$$ Hence $16m^2 - (8n-5)^2 = 231$, which factorises as $$\bigl(4m - (8n-5)\bigr)\bigl(4m + (8n-5)\bigr) = 231 = 3 \cdot 7 \cdot 11.$$

Write $u$ and $v$ for the two factors, so $uv = 231$, $u + v = 8m$ and $v - u = 2(8n-5)$. Since $m$ is a positive integer, $u + v > 0$, and as $uv > 0$ both are positive. The condition that really cuts the list down is $8 \mid u+v$; and once that holds, $v - u$ must also be twice an odd number of the right shape for $n$ to be an integer.

Running through the eight ordered factorisations: $$\begin{array}{c|c|c|c} (u,v) & u+v & m & 8n-5 = (v-u)/2 \\ \hline (1,231) & 232 & 29 & 115 \Rightarrow n = 15 \\ (3,77) & 80 & 10 & 37 \ \text{(not $\equiv 3 \bmod 8$)} \\ (7,33) & 40 & 5 & 13 \ \text{(no)} \\ (11,21) & 32 & 4 & 5 \ \text{(no)} \\ (21,11) & 32 & 4 & -5 \Rightarrow n = 0 \\ (33,7) & 40 & 5 & -13 \Rightarrow n = -1 \\ (77,3) & 80 & 10 & -37 \Rightarrow n = -4 \\ (231,1) & 232 & 29 & -115 \ \text{(no)} \end{array}$$ The surviving pairs $(m,n)$ are $(29,15)$, $(4,0)$, $(5,-1)$ and $(10,-4)$, with $|m-n|$ equal to $14$, $4$, $6$ and $14$. The maximum is $$|m - n| = 14.$$

The problem’s insistence that $n$ may have any sign is doing real work: three of the four solutions have $n\le0$, and one of them attains the maximum.

Answer 14

Solution: IOQM 2024, Q25

Let $M$ contain $k$ squares along with $92$. From the two averages, $$\text{sum of } M = 85(k+1), \qquad \text{sum of the squares} = 84k,$$ and these differ by $92$: $$85(k+1) - 84k = 92 \quad\Longrightarrow\quad k + 85 = 92 \quad\Longrightarrow\quad k = 7.$$ So there are seven distinct positive squares with total $84 \cdot 7 = 588$.

To make one square as large as possible, make the other six as small as possible. The six smallest distinct positive squares are $1, 4, 9, 16, 25, 36$, totalling $91$, so the largest square is at most $588 - 91 = 497$, giving $N \le 22$ since $22^2 = 484$ and $23^2 = 529$.

It remains to check that $N = 22$ is attainable, that is, that six distinct squares different from $484$ sum to $588 - 484 = 104$. Starting from the minimum $91$ we need $13$ more, and replacing $36$ by $49$ does exactly that: $$1 + 4 + 9 + 16 + 25 + 49 = 104.$$ So $M = \{1, 4, 9, 16, 25, 49, 484, 92\}$ works, and $N = 22$.

Answer 22

Solution: IOQM 2024, Q26

Put $n = \lfloor x \rfloor$ and $f(x) = 15x^2 + 15x + 16$, so the equation reads $f(x) = n^3$ with $x$ in the window $[n, n+1)$. First, $n$ is positive: $f(x) = 15\left(x + \tfrac12\right)^2 + \tfrac{49}{4} > 0$, so $n^3 > 0$. On non-negative arguments $f$ is increasing, since for $0 \le u < v$, $$f(v) - f(u) = 15(v-u)(v+u+1) > 0.$$ So the window contains a solution exactly when $$f(n) \le n^3 < f(n+1),$$ and both halves can be settled exactly.

The left inequality

Here $15n^2 + 15n + 16 \le n^3$, that is $n^3 - 15n^2 - 15n - 16 \ge 0$. This cubic factors: $$n^3 - 15n^2 - 15n - 16 = (n-16)\left(n^2 + n + 1\right),$$ and the quadratic factor is positive for every $n$, so the condition is exactly $n \ge 16$.

The right inequality

Here $n^3 < 15(n+1)^2 + 15(n+1) + 16 = 15n^2 + 45n + 46$, that is $g(n) < 0$ where $$g(n) = n^3 - 15n^2 - 45n - 46.$$ Now $g(17) = -233$ and $g(18) = 116$, so $17$ passes and $18$ fails. And $g$ increases from there on, since $$g(n+1) - g(n) = 3n^2 - 27n - 59,$$ which is positive at $n = 18$, where it equals $427$, and grows with $n$. So no $n \ge 18$ passes.

Putting the two together, $n = 16$ and $n = 17$ are the only values that work, and the required sum is $16 + 17 = 33$. The case $n = 16$ is worth seeing exactly: $60 \cdot 4096 - 735 = 495^2$, so $x = (495-15)/30 = 16$ lands on the left end of its own window.

Answer 33

Solution: IOQM 2024, Q30

Let the legs be $a = BC$ and $c = AB$ and the hypotenuse $b = AC$. The altitude to the hypotenuse satisfies $BD \cdot AC = AB \cdot BC$, so $$12b = ac.$$ Also $a^2 + c^2 = b^2$, and therefore $$(a+c)^2 = b^2 + 2ac = b^2 + 24b.$$ The perimeter is $a + c + b$, and $b$ is an integer, so the perimeter is an integer exactly when $a + c$ is, that is, exactly when $b^2 + 24b$ is a perfect square, say $k^2$.

Complete the square: $(b+12)^2 - k^2 = 144$, so $$(b + 12 - k)(b+12+k) = 144.$$ Both factors have the same parity and their product is even, so both are even; writing them as $2s$ and $2t$ gives $st = 36$ and $b + 12 = s + t$, that is $$b = s + t - 12, \qquad st = 36.$$

One more constraint decides the matter. Since $(a-c)^2\ge0$, we have $2ac\le a^2+c^2=b^2$. Together with $ac=12b$, this gives $$24b\le b^2,\qquad b\ge24,$$ as $b>0$. Running through the factorisations of $36$: $$\begin{array}{c|c|c} (s,t) & s+t & b \\ \hline (1,36) & 37 & 25 \\ (2,18) & 20 & 8 \\ (3,12) & 15 & 3 \\ (4,9) & 13 & 1 \\ (6,6) & 12 & 0 \end{array}$$ Only $b = 25$ clears the bound. A check confirms that it is genuine: $a + c = \sqrt{625+600} = 35$ and $ac = 300$, so $a$ and $c$ are the roots of $t^2 - 35t + 300$, namely $15$ and $20$. The familiar $15$-$20$-$25$ triangle has altitude $15 \cdot 20 / 25 = 12$ and perimeter $60$.

So the minimum possible length of $AC$ is $25$.

Answer 25

Solution: IOQM 2025 Part SEP, Q28

Since $b^2 = 4^a + 4a^2 + 4 > (2^a)^2$, we have $b > 2^a$. On the other side, $$(2^a+1)^2 = 4^a + 2^{a+1} + 1,$$ so $b < 2^a + 1$ as soon as $4a^2 + 4 < 2^{a+1}+1$, that is $$4a^2 + 3 < 2^{a+1}.$$ This holds for every $a \ge 7$, by induction. At $a = 7$ it reads $199 < 256$. And if it holds at some $a \ge 7$, then $$4(a+1)^2 + 3 < 2\left(4a^2+3\right) < 2 \cdot 2^{a+1} = 2^{a+2},$$ where the first step is the inequality $4a^2 + 8a + 7 < 8a^2 + 6$, which rearranges to $4a^2 - 8a - 1 > 0$ and is true for every $a \ge 3$.

So for $a \ge 7$ there is no integer strictly between $2^a$ and $2^a+1$, and no solution exists.

That leaves $a \le 6$, which is a short check: $$\begin{array}{c|c|c} a & 4^a+4a^2+4 & \text{square?} \\ \hline 1 & 12 & \text{no} \\ 2 & 36 & 6^2 \\ 3 & 104 & \text{no} \\ 4 & 324 & 18^2 \\ 5 & 1128 & \text{no} \\ 6 & 4244 & \text{no} \end{array}$$ The solutions are $(a,b) = (2,6)$ and $(4,18)$, and the larger value of $a+b$ is $$4 + 18 = 22.$$

Answer 22

Solution: IOQM 2026, Q15

Write $s = a + b$ and $t = c + d$, so $t \ge 2$ and the total is $s + t$.

First, $s \ge 47$. Since $(a - 1)(b - 1) \ge 0$, we have $ab \ge s - 1$, and so $$a^2 + b^2 = s^2 - 2ab \le s^2 - 2s + 2 = (s - 1)^2 + 1.$$ If $s \le 46$ this is at most $45^2 + 1 = 2026$, but $a^2 + b^2 = 2026 + cd^2$ is at least $2027$.

Then parity. A square has the same parity as the number squared, so $a^2 + b^2$ has the parity of $s$, while $2026 + cd^2$ has the parity of $cd$. A total of at most $50$ with $s \ge 47$ and $t \ge 2$ leaves two cases.

• $s = 47$ and $t \le 3$. Here $s$ is odd, so $cd$ is odd, so $c$ and $d$ are both odd and $t = 2$, with $c = d = 1$. Then $a^2 + (47 - a)^2 = 2027$, which simplifies to $a^2 - 47a + 91 = 0$. Its discriminant is $47^2 - 364 = 1845$, which lies strictly between $42^2 = 1764$ and $43^2 = 1849$, so there is no integer $a$.

• $s = 48$ and $t = 2$. Then $c = d = 1$ and $cd$ is odd while $s$ is even, which parity forbids.

So the total is at least $51$, and it is reached: $a = 45$, $b = 2$, $c = 3$, $d = 1$ gives $2025 + 4 - 3 = 2026$ and $a + b + c + d = 51$.

Answer 51

Solution: IOQM 2026, Q28

Doubling nine times, $b_{10} = 512\,b_1$, and $512\,b_1 < 2026$ leaves $b_1 = 1, 2, 3$.

Running the rule for $a$ forward from $a_1$ and $a_2$, $$\begin{array}{c|l} n&a_n\\\hline 3&a_1+a_2\\ 4&a_1+2a_2\\ 5&2a_1+3a_2\\ 6&3a_1+5a_2\\ 7&5a_1+8a_2\\ 8&8a_1+13a_2\\ 9&13a_1+21a_2\\ 10&21a_1+34a_2 \end{array}$$ the coefficients being consecutive Fibonacci numbers. The sequence increases exactly when $a_1 < a_2$, since every later term is a sum of two earlier positive ones.

Write $k = b_1$, so the equation is $21a_1 + 34a_2 = 512k$. Since $546 = 26 \times 21$, $$34a_2 - 512k = 34(a_2 + k) - 546k$$ is a multiple of $21$ exactly when $34(a_2 + k)$ is, and as $34$ and $21$ share no factor, $a_2 + k$ must be a multiple of $21$. Also $34a_2 < 512k$.

• $k = 1$: $a_2 \le 15$, and $a_2 + 1$ is never a multiple of $21$. No solution.

• $k = 2$: $a_2 \le 30$ and $a_2 + 2$ a multiple of $21$ give $a_2 = 19$, then $a_1 = 18$. Since $18 < 19$ this works, and $a_1 + b_1 = 20$.

• $k = 3$: $a_2 \le 45$ gives $a_2 = 18$ or $39$. The first needs $a_1 = 44$, which is not less than $18$; the second gives $a_1 = 10$, which works, and $a_1 + b_1 = 13$.

The possible values are $20$ and $13$, and their sum is $33$.

Answer 33

Solution: IOQM 2025 Part SEP, Q7

Squaring the given equation, $$x - \sqrt{x+a} = a - 2y\sqrt a + y^2,$$ that is, $$x - a - y^2 = \sqrt{x+a} - 2y\sqrt a.$$ The left-hand side is an integer, so $\sqrt{x+a} = m + 2y\sqrt a$ for the integer $m = x - a - y^2$. Squaring again, $$x + a = m^2 + 4y^2a + 4my\sqrt a,$$ and since $a$ is not a perfect square, $\sqrt a$ is irrational, so $4my = 0$.

If $m = 0$ then $\sqrt{x+a} = 2y\sqrt a$, hence $x = (4y^2-1)a$ and $m = x - a - y^2 = (4y^2 - 2)a - y^2 = 0$, giving $a = \dfrac{y^2}{4y^2-2} < 1$, which no positive integer $a$ satisfies. So $y = 0$.

With $y = 0$ the original equation reads $x - \sqrt{x+a} = a$, so $\sqrt{x+a} = x - a$. Setting $t = x - a \ge 0$ and squaring, $x + a = t^2$, and subtracting the two expressions for $x$, $$t^2 - 2a = t, \qquad a = \frac{t(t-1)}{2}.$$ So $a$ is a triangular number. Conversely, every triangular number $a = \tfrac12 t(t-1)$ gives a solution, taking $y = 0$ and $x = t^2 - a$; the hypothesis that $a$ is not a perfect square then keeps only those triangular numbers that are not squares, which rules out $1$ and $36$ among the small ones.

The largest triangular number below $100$ is $\tfrac{14 \cdot 13}{2} = 91$, and $91$ is not a perfect square, as required. (Here $t = 14$ and $x = 196 - 91 = 105$.)

Answer 91