# Prove that there do not exist positive integers $x$ and $y$ with $x^2 - y^2 = n$

I'm working on a homework problem that is as follows:

Suppose that $n$ is a positive even integer with $n/2$ odd. Prove that there do not exist positive integers $x$ and $y$ with $x^2 - y^2 = n$.

It looked like a good candidate for proof by contradiction. So I know that I still assume the argument "$n$ is a positive even integer with $n/2$ odd" but will try to show that there exists positive integers $x$ and $y$ with $x^2 - y^2 = n$, and if this reaches a contradiction then I have proven the original conjecture.

So I started with the $n/2$ is odd and rewrote it as $n/2 = 2k+1$, for some $k$ in the integers.

Then I knew that $n = 4k+2$, and then tried so equate that with $x^2 - y^2 = 4k+2$.

EDIT: I then recognized that $x^2 - y^2$ is equivalent to $(x + y)(x - y)$ but that doesn't seem to be very helpful, because if you divide one or the other out you get a term on the RHS in terms of k and x and y.

I'm going to keep playing with it but I don't really have any good strategies going forward. Any help is appreciated!

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$x^2-y^2=(x-y)(x+y)$ – Chris Eagle Jul 29 '12 at 21:49
$(x-y)(x+y)$ - both even or both odd. – Old John Jul 29 '12 at 21:51
Also, $x^2$ is even if and only if $x$ is even, same for $y.$ Oh: the first thing is that your $n$ is not divisible by 4. – Will Jagy Jul 29 '12 at 21:51
Dear Augustus, If you want to argue by contradiction, you do not "try to show that there exists positive integers $x$ and $y$ with ...", but rather you assume that such $x$ and $y$ exist. Regards, – Matt E Jul 30 '12 at 3:34
My bad Matt E. slip of the tongue. Thanks greatly for the correction – Arthur Collé Jul 30 '12 at 4:07

A more or less mechanical approach is to work modulo $4$. Note that for any integer $k$, $k^2\equiv 0 \pmod{4}$ or $k^2\equiv 1\pmod{4}$.

So, modulo $4$, $x^2-y^2$ can only take on the values $0$, $1$, and $-1$.

More basic, and more useful, is to suppose that $x^2-y^2=n$. Then $(x-y)(x+y)=n$. Note that for any integers $x$ and $y$, the numbers $x-y$ and $x+y$ are both even or both odd. (If we want a proof, their difference $2y$ is even.)

In neither case is $(x-y)(x+y)$ twice an odd integer. You started along these lines. Note that you were one step from the end.

Remark: The reason the second idea is more useful is that when it comes to solving $x^2-y^2=n$, we express $n$ as a product $st$ of integers of the same parity, and solve the system $x-y=s$, $x+y=t$. The solution is $x=\frac{s+t}{2}$, $y=\frac{s-t}{2}$. If $n$ is twice an odd integer, then this process breaks down, because one of $s$ and $t$ will be odd and the other even, so we do not get integers $x$ and $y$.

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I like this a lot. +1 – mixedmath Jul 29 '12 at 21:54
I don't understand the congruence that you have written for the first part before the line break. How is it that you arrive at 4 divides $k^2$ or 4 divides $k^2 - 1$? – Arthur Collé Jul 29 '12 at 22:06
@Augustus: You might just calculate it - if you have integers of the form $4k, 4k+1, 4k+2, 4k+3$ and square them, they'll fall into the two residue classes $0\pmod 4$ or $1 \pmod 4$. Alternately, if you're familiar with quadratic residues, then you could use that approach. – mixedmath Jul 29 '12 at 22:11
If $k$ is even, say $k=2q$, then $k^2=4q^2$, so $4$ divides $k^2$. If $k$ is even, say $k=2q+1$. then $k^2=(2q+1)^2=4q^2+4q+1=4(q^2+q)+1$, which is $1$ more than a multiple of $4$. Actually, $q^2+q$ is even, so indeed $k^2$ is $1$ more than a multiple of $8$, an often useful fact. – André Nicolas Jul 29 '12 at 22:11
Okay thanks alot! – Arthur Collé Jul 29 '12 at 22:14

HINT

1. $x^2 - y^2 = (x-y)(x+y)$
2. If $x-y$ is even, then so is $x+y$
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For every integer $x$, $x^2=0$ or $1\pmod{4}$. Hence, for every integers $x$ and $y$, $x^2-y^2=1$ or $0$ or $-1\pmod{4}$. On the other hand, if $n=2(2k+1)$ for some integer $k$, $n=2\pmod{4}$. Hence, $x^2-y^2=n\pmod{4}$ with $n$ twice an odd integer is impossible.

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$\rm\begin{eqnarray}{\bf Hint}\,\ \ \ If\ prime\,\ p\:|\:a\!-\!b\,\ then\,\ p\:|\:ab\:&\Rightarrow&\:\rm p^2\:|\:ab \\ \rm Therefore\ we\ deduce\ that\,\ 2\:|\:x^2\!-\!y^2&\Rightarrow&\:\rm\ \ 4\:|\:x^2\!-\!y^2\ \ for\ \ p=2,\,\ a,b = x\pm y\end{eqnarray}$

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If $x^2-y^2= N$ and $N/2$ is odd this diophantine equation has no solutions. In fact you can rewrite this equation: $(x-y)(x+y)=N$, but $N$ is even and it is divisibile for $2$ then only one factor is even. So solving this equation with this formula: $x=(p+q)/2$ and $y=(p-q)/2$, where $pq=N$ you obtain two rational solutions.

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