$x^2-y^2=2s$, s cannot be an odd integer How can we prove that if $x^2-y^2=2s$ holds, s cannot be an odd integer. What theorem in number theory should we use?
 A: Since $x^2-y^2=(x+y)(x-y)$, if $x^2-y^2$ is even, then $x+y$ is even or $x-y$ is even. But if one of them is even, so is the other. This implies that their product  is a multiple of $4$ and so is not twice an odd number.
In case it's not clear, the theorems used in this argument are:


*

*the product of two odd numbers is odd.

*the sum of two even numbers is even.
A: The remainder of a square when divided by $4$ can be $0$ or $1$. Hence the remainder when divided by $4$ of a difference of squares can be $0,1,3$. If it is even it can only be $0$. So an even difference of squares is a multiple of $4$.
A: $x^2-y^2=(x+y)(x-y)$
Let $y=x+a$ then
$(x+y)(x-y)=(2x+a)(2x-a)=4x^2-a^2$
If $4x^2-a^2=2s$ then $a^2$ must be divisible by 2. Since $a$ is an integer $a$ must be divisible by 2. Thus $a=2b$. Thus $2s=4x^2-4b^2=4c$ where $c=x^2-b^2$ is an integer. Thus $s$ is divisible by $2$.
A: Direct proof by contradiction works.
Notice that either both $x$ and $y$ must be even, or both must be odd, as the RHS is even.
If $x$ and $y$ are both even, then the LHS is divisible by $4$, and hence $4$ divides the RHS, ans so $s$ has to be even.
If $x$ and $y$ are both odd, then they are of the form $4k+1$ or $4k+3$. In either case, their square is of the form $4n+1$, and hence the LHS is divisible by $4$.
A: You don't need any special theorem, just reason through it:
If $x^2 - y^2$ is even, that means either both $x$ and $y$ are even or they're both odd.
If both $x$ and $y$ are even, let's say $x = 2m$ and $y = 2n$. Then $x^2 - y^2 = 4m^2 - 4n^2 = 4(m^2 - n^2)$ and $s = 2(m^2 - n^2)$, meaning that it's even.
If both $x$ and $y$ are odd, let's say $x = 2m + 1$ and $y = 2n + 1$. Then $x^2 - y^2 = (4m^2 + 4m + 1) - (4n^2 + 4n + 1) = 4(m^2 + m - n^2 - n)$ and $s = 2(m^2 + m - n^2 - n)$, meaning that it's even again.
