Prove or disprove that every even integer is the difference of two squares. I don't know where to start. Let $2k$ be an even integers and $a^2 - b^2$...
I do not believe this question is correct as stated. A counterexample is the number $2$. $2$ is not the difference of squares. If it were, say $a^2 - b^2 = 2$, then $(a-b)$ would divide $2$. So $(a-b)$ must one of $1$ or $2$, since $a-b$ is positive. $a+b$ must then be $2$ or $1$ respectively. In either case, we are forced to have that $(a+b) + (a-b) = 2a = 3$. But $2$ does not divide $3$. 

What is true, however, is that every integer of the form $4k$ for some integer $k$ is the difference of squares. We may assume $k$ is positive. If $k = 0$, then $a=b=0$ works. If $k$ is negative, then assuming the above, there are $a$ and $b$ so that $a^2 - b^2 = -k$. Then $b^2 - a^2 = k$. 

Now, simply notice that $$(k+1)^2 - (k-1)^2 = k^2 + 2k + 1 - k^2 + 2k - 1 = 4k,$$
so $4k$ is the difference of squares.
I think we are lacking informations...

Trivial in R
If a and b belong to R then if if k positive a = sqrt(2k) and b = 0. If k negative a = 0 b = sqrt (-2k)

False in N
Starting point is the fact that every odd number is the difference if two squares.