# Algebra

#### My solutions to the problems in the excellent book on Problem Solving by Loren Larson.

By Vamshi Jandhyala in mathematics problem solving

April 22, 2020

## Problem 4.1.5

a. If $a$ and $b$ are consecutive integers, show that $a^2+b^2+(ab)^2$ is a perfect square.

b. If $2a$ is the harmonic mean of $b$ and $c$, show that the sum of the squares of the three number $a,b,$ and c is the square of a rational number.

c. If $N$ differs from two successive squares between which it lies by $x$ and $y$ respectively, prove that $N-xy$ is a square.

### Solution

a. If $a=n$ and $b=n+1$, we have

$$
\begin{align*}
a^2+b^2+(ab)^2 &= n^2 + (n+1)^2 + n^2(n+1)^2 \\

&= n^4 + 2n^3 + 3n^2 + 2n + 1 \\

&= (n^2+n+1)^2
\end{align*}
$$

b. We have

$$
\begin{align*}
a^2+b^2+c^2 &= \frac{b^2c^2}{(b+c)^2} + b^2 + c^2 \\

&= \frac{b^4+c^4+3b^2c^2+2b^3c+2c^3b}{(b+c)^2} \\

&= \left(\frac{b^2+c^2+bc}{b+c}\right)^2
\end{align*}
$$

c. Let $x=N-n^2$ and $y=(n+1)^2-N$, we have

$$
\begin{align*}
N-xy &=N-(N-n^2)((n+1)^2-N)\\

&= n^2(n+1)^2 +N^2 + N(1 - (n+1)^2-n^2)\\

&= (N - n(n+1))^2
\end{align*}
$$

## Problem {4.1.7}

Suppose that an integer $n$ is the sum of two triangular numbers,

$$ \begin{align*} n= \frac{a^2+a}{2} + \frac{b^2+b}{2} \end{align*} $$

write $4n+1$ as the sum of two squares, $4n+1=x^2+y^2$, and show how $x$ and $y$ can be expressed in terms of $a$ and $b$.

Show that conversely, if $4n+1=x^2+y^2$, the $n$ is the sum of two triangular numbers.

### Solution

We have,

$$
\begin{align*}
4n+1 &= 2(a^2+a)+2(b^2+b)+1\\

&= (a+b+1)^2 + (a-b)^2
\end{align*}
$$

If $4n+1=x^2+y^2$, we have

$$
\begin{align*}
n &= \frac{x^2+y^2-1}{4} \\

&= \frac{(x+y-1)}{2}\frac{(x+y+1)}{2}/2 + \frac{(x-y-1)}{2}\frac{(x-y+1)}{2}/2
\end{align*}
$$

## Problem 4.1.8

Let $N$ be the number which when expressed in decimal notation consists of $91$ ones:

$$ \begin{align*} N = \underbrace{111\dots1}_{91}. \end{align*} $$

Show that $N$ is a composite number.

### Solution

We have

$$ \begin{align*} N = \frac{10^{91}-1}{10-1} = \frac{(10^7)^{13}-1}{10-1} \end{align*} $$

As $a^n-b^n$ is divisible by $a-b$, the numerator is divisible by $10^7-1=9\cdot1111111$. As $N$ is divisible by $1111111$, $N$ is a composite number.

## Problem 4.2.8

What is the greatest common divisor of $x^n-1$ and $x^m-1$?

### Solution

The greatest common divisor of $x^n-1$ and $x^m-1$ is $x^{gcd(m,n)}-1$ as $x^n-1$ is divisible $x^k-1$ whenever $k$ divides $n$.

## Problem 4.2.9

Let $f(x)$ be a polynomial leaving a remained $A$ when divided by $x-a$ and the remainder $B$ when divided $x-b$, $a\neq b$. Find the remainder when $f(x)$ is divided by $(x-a)(x-b)$.

### Solution

We have

$$
\begin{align*}
f(x) = q(x)(x-a)(x-b) + cx+d \\

\implies f(a) = A = ca+d \wedge f(b) = B = cb+d
\end{align*}
$$

Solving for $c$ and $d$ we get the remainder when $f(x)$ is divided by $(x-a)(x-b)$ as

$$ \begin{align*} \frac{A-B}{a-b}x + \frac{bA-aB}{b-a} \end{align*} $$

## Problem 4.2.10

Show that $x^{4a}+x^{4b+1}+x^{4c+2}+x^{4d+3}$,a,b,c,d positive integers is divisible by $x^3+x^2+x+1$.

### Solution

Let $f(x)= x^{4a}+x^{4b+1}+x^{4c+2}+x^{4d+3}$.

We have $f(-1)=0,f(i)=0$ and $f(-i)=0$, therefore by the Factor theorem,we see that $(x+1)$, $(x-i)$ and $(x+i)$ are factors of $f(x)$. Therefore $f(x)$ is divisible by $(x+1)(x-i)(x+i)=1+x+x^2+x^3$.

## Problem 4.2.11

Show that the polynomials $(cos\theta+xsin\theta)^n-cos(n\theta)-xsin(n\theta)$ is divisible by $x^2+1$.

### Solution

Let $f(x)= (cos\theta+xsin\theta)^n-cos(n\theta)-xsin(n\theta)$.We have $f(i)=0$ and $f(-i)=0$ using De Moivre’s Theorem. Using the Factor theorem,we get that $(x+i)$ and $(x-i)$ are factors of $f(x)$. Therefore $f(x)$ is divisible by $(x-i)(x+i)=1+x^2$.

## Problem 4.2.12

For what $n$ is the polynomial $1+x^2+x^4+\dots+x^{2n-2}$ divisible by the polynomial $1+x^2+x^3+\dots+x^{n-1}$.

### Solution

We have

$$
\begin{align*}
1+x^2+x^4+\dots+x^{2n-2} &= \frac{x^{2n}-1}{x^2-1} \\

1+x^2+x^3+\dots+x^{n-1} &= \frac{x^{2n}-1}{x^2-1}
\end{align*}
$$

Dividing the first expression by the second we get

$$ \begin{align*} \frac{(x^{2n}-1)(x-1)}{(x^2-1)(x^n-1)} = \frac{x^n+1}{x+1} \end{align*} $$

$x^n+1$ is divisible by $x+1$ when $n$ is odd, therefore the polynomial $1+x^2+x^4+\dots+x^{2n-2}$ divisible by the polynomial $1+x^2+x^3+\dots+x^{n-1}$ when $n$ is odd.

## Problem 4.2.13

A real number is called \emph{algebraic} if it is a zero of a polynomial with integer coefficients.

a. Show that $\sqrt{2}+\sqrt{3}$ is algebraic.

### Solution

a. We have

$$
\begin{align*}
x - \sqrt{2} - \sqrt{3} &= 0\\

\implies x^2 + 2 -2\sqrt{2}x &= 3 \\

\implies x^4-2x^2+1 &= 8x^2 \\

\implies x^4 -10x^2+1 &= 0
\end{align*}
$$

As $\sqrt{2}+\sqrt{3}$ is a root of the equation $ x^4 -10x^2+1 = 0$ with integer coefficients it is algebraic.

## Problem 4.2.17

a. Suppose $f(x)$ is a polynomial over the real numbers and $g(x)$ is a divisor of $f(x)$ and $f'(x)$. Show that $(g(x))^2$ divides $f(x)$.

b. Use the idea of part (a) to factor $x^6+x^4+3x^2+2x+2$ into a product of irreducibles over the complex numbers.

### Solution

a. We have $f(x)=g(x)q(x)$. From this we have, $f'(x)=g(x)q'(x)+g'(x)q(x)$. As $g(x)$ is also a divisor of $f'(x)$, we can easily see that $q(x)$ should be divisible by $g(x)$ as $g(x)$ cannot divide $g'(x)$ which is a polynomial of lower degree than $g(x)$. Therefore, $(g(x))^2$ divides $f(x)$.

b. Let $f(x)=x^6+x^4+3x^2+2x+2$, we have $f'(x)=6x^5+4x^3+6x+2$.

## Problem 4.2.18

Determine all pairs of positive integers $(m,n)$ such that $1+x^n+x^{2n}+\cdots+x^{mn}$ is divisible by $1+x+x^2+\dots+x^m$.

### Solution

Denote the first and larger polynomial to be $f(x)$ and the second one to be $g(x)$. We could instead consider $f(x)$ modulo $g(x)$. Notice that $x^{m+1} = 1 \pmod {g(x)}$, and thus we can reduce the exponents of $f(x)$ to their equivalent modulo $m+1$. We want the resulting $h(x)$ with degree less than $m+1$ to be equal to $g(x)$ (of degree $m$), which implies that the exponents of $f(x)$ must be all different modulo $m+1$. This can only occur if and only if $gcd(m+1, n) = 1$.

## Problem 4.3.13

Show that

$$
\begin{align*}
(1+x)^n-x(1+x)^n + x^2(1+x)^n - \dots x^k(1+x)^n \\

= (1+x)^{n-1} (1-(-x)^{k+1}),
\end{align*}
$$

and use this identity to prove that

$$ \begin{align*} {n-1 \choose k}= {n \choose k}-{n \choose k-1}+\cdots \pm {n \choose 0} \end{align*} $$

### Solution

We have

$$
\begin{align*}
(1+x)^n-x(1+x)^n + x^2(1+x)^n - \dots x^k(1+x)^n \\

= (1+x)^n(1-x+x^2-\dots\pm x^k) \\

= (1+x)^n \frac{(1 - (-x)^{-k+1})}{1-(-x)} \\

= (1+x)^{n-1} (1-(-x)^{k+1})
\end{align*}
$$

The coefficient of $x^k$ on the left hand side is

$$ \begin{align*} {n \choose k}-{n \choose k-1}+\cdots \pm {n \choose 0} \end{align*} $$

as we only need the term $x^{k-i}$ from $(1+x)^n$ from the $i^{th}$ term.

The coefficient of $x^k$ on the right hand side is

$$ \begin{align*} {n-1 \choose k} \end{align*} $$

## Problem 4.3.19

a. Solve the equation $x^3-3x^2+4=0$, given that two of its roots are equal.

b. Solve the equation $x^3-9x^2+23x-15=0$, given that its roots are in arithmetical progression.

### Solution

a. Let $f(x)= x^3-3x^2+4$. As $f(x)=0$ has two equal roots, $f(x)=0$ and $f'(x)=0$ share that root. The roots of $f'(x)=3x^2-6x=0$ are $x=0$ and $x=2$. As $x=0$ is not a root of $f(x)=0$, $x=2$ is a common root of $f(x)$ and $f'(x)$.As the product of the roots of $f(x)=0$ is $-4$, the roots of $f(x)=0$ are $-1,2$ and $2$.

b. Let $a-d,a,a+d$ be the roots of $x^3-9x^2+23x-15=0$. We have sum of the roots is $3a=9$ and the product of the roots is $3(9-d^2)=15$. Therefore, the roots of the equation are $1,3$ and $5$.

## Problem 4.3.20

Given $r,s,t$ are the roots of $x^3+ax^2+bx+c=0$,

a. Evaluate $1/r^2+1/s^2+1/t^2$, provided that $c \neq 0$.

b. Find a polynomial whose roots are $r^2,s^2,t^2$

### Solution

a. Let $f(x)=x^3+ax^2+bx+c$.As $c \neq 0$, none of the roots of $f(x)=0$ can be zero.We now need to find $g$ such that $y=1/z^2$ is a root of $g(y)=0$ where $z$ is the root of $f(x)=0$.Therefore, $1/r^2+1/s^2+1/t^2$ will be the sum of the roots of $g(y)=0$.We have $f(1/\sqrt{y})=0$, which means

$$
\begin{align*}
(\frac{1}{\sqrt{y}})^3 + a(\frac{1}{\sqrt{y}})^2+ b(\frac{1}{\sqrt{y}}) + c &= 0 \\

\implies c^2y^3 +y^2(2ac-b^2)+y(a^2-2b)-1 &=0
\end{align*}
$$

Therefore, $1/r^2+1/s^2+1/t^2 = b^2-2ac$.

b. We now need to find $h(y)$ such that $y=z^2$ is a root of $h(y)=0$ where $z$ is the root of $f(x)=0$. We have $f(\sqrt{y})=0$, which means

$$
\begin{align*}
\sqrt{y}^3 + a\sqrt{y}^2+ b\sqrt{y} + c &= 0 \\

\implies y^3 +y^2(2b-a^2)+y(b^2-2ac)-c^2 &=0
\end{align*}
$$