Exercise 5.3.7 ( For which integers $n$ are there solutions to the equation $x^2 - y^2 =n$?)

Proof. Let $n\in \mathbb{Z}$ be any integer such that $n=x^2-y^2$ for suitable integers $x,y$. Then $n=(x-y)(x+y)$. We write $d=x-y,e=x+y$, so that $d,e$ are divisors of $n$ such that

thus $e,d$ have same parity (since $d-e=2y$), and

This implies that $n$ has some divisor $d$ such that $d$ and $n∕d$ have same parity. This is the case if $n$ is odd, since $n$ and $1$ are both odd. This is also the case if $n\equiv 0(\mathit{mod}4)$, because $2$ and $n∕2$ are even.

It remains only the case $n\equiv 2(\mathit{mod}4)$, so $n=2(2k+1)$ for some integer $k$. Let $d$ be any divisor of $n$. If $d$ is odd, then $n∕d$ is even, and if $d=2d^{\prime }$ is even, then $n∕d=(2k+1)∕d^{\prime }$ is odd. So no divisor $d$ of $n$ is such that $d$ and $n∕d$ have same parity. This shows that in this case $n=x^2-y^2$ has no solution.

(More concisely, we note that $x^2\equiv 0$ or $1(\mathit{mod}4)$, thus $x^2-y^2\equiv 0,1$ or $-1(\mathit{mod}4)$, so $n=x^2-y^2\not\equiv 2(\mathit{mod}4)$.)

In the other cases,

• If $n\equiv 1(\mathit{mod}2)$, then

  $$n={\left(\frac{n+1}{2}\right)}^2-{\left(\frac{n-1}{2}\right)}^2,$$

  where $x=\frac{n+1}{2},y=\frac{n-1}{2}$ are integers.

• If $n\equiv 0(\mathit{mod}4)$, then

  $$n={\left(\frac{n}{4}+1\right)}^2-{\left(\frac{n}{4}-1\right)}^2,$$

  where $x=\frac{n}{4}+1,y=\frac{n}{4}-1$ are integers.

In both cases $n=x^2-y^2$ has integer solutions.

In conclusion, the equation $x^2-y^2=n$ has solutions if and only if $n\not\equiv 2(\mathit{mod}4)$. □