Prove that if $a$ and $b$ are some integers, then $a^2 - 4b$ does not equal $2$.

Question

Prove that if $a$ and $b$ are some integers, then $a^2 - 4b$ does not equal $2$.

How do I prove this, I tried using contradiction but I'm not sure how I can contradict this statement?

Answer 1

Hint:

Use congruences. Such an equality would imply $2$ is a square modulo $4$. Write the list of all squares modulo $4$ and see if $2$ is among them.

Some details:

Remember two integers are congruent modulo $4$ if they have the same remainder when divided by $4$. This is equivalent to ‘one of them is equal to the other, pus a multiple of $4$’.

So $a^2-4b=2\iff a^2=2+4b\iff a^2\equiv 2\mod 4$.

Now the classes of congruence modulo $4$ are $0$, $\pm1$, $2$, and the relation of congruence is compatible with addition and multiplication. Thus the clsses of congruence of the squares are

• $0^2=0$;
• $(\pm1)^2=1$;
• $2^2=4\equiv0$.

Neither $-1$ nor $2$ are in the list of (congruence classes) of squares modulo 4: we proved $2$ and $-1$ are not squares modulo $4$.

Answer 2

Your inequality can be rewritten as $a^2\neq 4b+2= 2(2b+1)$. Note that $2b+1$ is always an odd number whatever be $b$. SO your question boils down to showing $a^2$ can never be double an odd number.

When $a$ is odd its square cannot be double of any number, being odd again.

When $a$ is even $ a^2$ will be an even number squared, that is $2m\times 2m$ which is double of an even number. Done.