# 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?

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If you do not have a good idea, just check all cases: a even, b even; a even, b odd; a odd, b even; a odd, b odd. (There is some theory behind the fact that this kind of proof has to work, but you do not even have to know that.) – Carsten S Nov 1 '13 at 16:50
Modulo 2, $a^2=a,b^2=b$, $-5=1$; so you can say $a^2-5b=b^2-5a$. – Pedro Tamaroff Nov 1 '13 at 17:36

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:

1. $a$ is even; and
2. $a$ 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.

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I'm sorry - i don't quite follow; when you say 5b=a^2−(a^2−5b) is the difference of two even numbers... How is this the case? I don't understand :/ – Omar Qassem Dec 2 '12 at 1:56
@Omar: That $5b=a^2-\left(a^2-5b\right)$ is just algebra. We assumed at the beginning that $a^2-5b$ is even, and in that paragraph we’re considering the case in which $a$ is even, which implies that $a^2$ is even. Thus, in $a^2-\left(a^2-5b\right)$ we’re subtracting one even number from another, and $5b$ is therefore the difference of two even numbers, one even number minus another even number. – Brian M. Scott Dec 2 '12 at 1:59
Ah that's amazing - thank you so much for your help :) – Omar Qassem Dec 2 '12 at 2:07
@Omar: You’re very welcome. – Brian M. Scott Dec 2 '12 at 2:10

If $a^2-5b$ is even then $a^2$ and $5b$ are both even or both odd.

In the first case, $a$ and $b$ are both even, so that $b^2$ and $5a$ are both even. In the second case, $a$ and $b$ are both odd, so that $b^2$ and $5a$ are both odd.

The result follows.

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