Showing that $b^2−5a$ is even if $a^2−5b$ is even Suppose $a$ and $b$ are integers and $a^2−5b$ is even. How can we prove that $b^2−5a$ is even?  

I’m going to assume that you mean $a^2-5b$ and $b^2-5a$. 
HINTS: Suppose that $a^2-5b$ is even, and consider two possibilities:


*

*$a$ is even; and  

*$a$ is odd.


If $a$ is even, then $a^2$ is even (why?). Then $5b=a^2-\left(a^2-5b\right)$ is the difference of two even numbers, so it must be even (why?), so $b$ is which, odd, or even?
If $a$ is odd, then $a^2$ is odd (why?), and you should again be able to say whether $5b$ is odd or even, and then whether $b$ is odd or even.
In both cases you’ll find that you know the parity (oddness or evenness) of both $a$ and $b$, and from that it’s possible to compute that $b^2-5a$ is indeed even.
