contradiction proof on divides

Question

Suppose a,b ∈ Z. If 4 | $(a^2 + b^2)$ then a and b are not both odd.

So, assuming that 4 | $(a^2 + b^2)$ and $a$ and $b$ are odd

this gives $4k=(2l+1)^2+(2u+1)^2$ for some $k,l,u\in z$

eventually leading to $4k=4(l^2+l+u)+2(u^2+1)$

The RHS is not a multiple of 4 when $u=2$ contradiction.

Is this valid, thanks.

Answer 1

It's not yet valid, because you haven't shown why the RHS cannot be a multiple of $4$. You cannot simply set $u=2$, because the $u$ you have is already determined by $b$, since $b=2u+1$.

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To correct your proof, re-think how you got from

$$4k=(2l+1)^2 + (2u+1)^2$$

to

$$4k = 4(l^2+l+u) + 2(u^2+1)$$

because I think you were a bit sloppy here.

Answer 2

This is invalid. You have done a mistake in calculation. RHS will be $4(l^2+u^2+l+u)+2$, which is not divisible by 4. Hope it helps:)

Answer 3

If you multiply correctly, the RHS leaves a rest of 2 modulo 4. So your idea was correct, you just need to recalculate.