Where is the Contradiction? Elementary Number Theory Proof : No natural numbers x, y such that $x^2 - y^2 = 2s$ where s odd integer.

Question

Let $m$ be a positive integer of the form $m = 2s$, where $s$ is an odd integer. Prove that there do not exist positive integers $x$ and $y$ such that $x^2 - y^2 = m.$

Proclaimed Solution via Proof by Contradiction: Assume, to the contrary, that there exist positive integers $x$ and $y$ such that $(x - y)(x + y)= m = 2s \quad \text{ where $s$ is an odd integer.} \tag{*}$ We consider two cases, according to whether $x$ and $y$ are of the same parity or of opposite parity. $\boxed{\text{Case 1:}}$ If $x$ and $y$ are of the same parity, then both $x - y$ and $x + y$ are even.
$\boxed{\text{Case 2:}}$ If $x$ and $y$ are of opposite parity, then both $x - y$ and $x + y$ are odd. Produce a contradiction in each case.

$\Large{1.}$ The product of two even numbers is always even. Thus, $(x - y)(x + y)$ is even. Then $(*)$ becomes: $\text{even = 2(odd)}$. Where is the contradiction?

The analogous solution works for Case 2? The product of two odds is always odd. Then $(*)$ becomes: $\text{odd = ($2s$ = even)}$. Contradiction.

$\Large{2.}$ What is the motivation behind considering "whether $x$ and $y$ are of the same parity or of opposite parity"?

Source: Exercise 5.22, P124 of Mathematical Proofs, 2nd ed. by Chartrand et al

Answer 1

1. Suppose that $x+y$ and $x-y$ are both even; then there are integers $k$ and $\ell$ such that $x+y=2k$ and $x-y=2\ell$. But then $2s= (x+y)(x-y)=4k\ell=2(2k\ell)$, so $m=2k\ell$, and $s$ is therefore even, contradicting the hypothesis that $s$ is odd.

2. The two cases need to be considered separately, because each requires a different argument. Are you also asking how you might realize that if confronted with the problem? That’s largely a matter experience and partly a matter of recognizing that when the hypotheses specifically involve parity ($m$ is odd), parity may play a rôle in the proof.

Answer 2

Regarding case 1: If $x - y$ and $x + y$ are both even, then $(x - y)(x + y)$ is divisible by $4$, and so $s$ cannot be odd.

As far as motivation, the technique is somewhat suggested by considering that the right hand side is divisible by $2$ exactly once - so the left side must be divisible by $2$ exactly once, so $x - y$ and $x + y$ must have opposite parity.

Answer 3

For the first case, the contradiction lies in the fact that both $\left(x-y\right)$ and $\left(x+y\right)$ are even. This means that \begin{eqnarray*} \left(x-y\right)\left(x+y\right) & = & \left(2x\right)\left(2y\right)\\ & = & 4xy. \end{eqnarray*} If a number is $2s=2\left(2k+1\right)=4k+2$, then it is not a multiple of four. That's the contradiction.

Answer 4

1) The contradiction is that $(x-y)(x+y)$ is divisible by $4$, but $2$(odd) is not.

2) The motivation is that it exhausts all possibilities. If x and y are not of the same parity, we have the contradiction that $(x+y)(x-y)$ is odd, but $2s$ isn't.