Proof that if equation is multiple of number, similar equation is multiple too

Question

Given $a$ and $b$ integers such that $a+5b$ and $5a-b$ are both multiples of 2002, prove that $a²+b²$ is multiple of 2002 as well.

Here is what I have:

Lemma 1: For all multiple $a$ and $b$ of $k$, $ab$ is multiple of $k$ too.

Proof: Let $n$ and $m$ be multiples of $p$; $(np)(mp)=p(mp)$

$(a+5b)(5a-b)=(5a²+24ab-5b²)$

I'm stuck from here, I tried dividing $(5a²+24ab-5b²)$ by $a²+b²$ thinking that if they are both multiples of 2002 they would give an exact expression. How to continue?

Answer 1

HINT: Brahmagupta-Fibonacci Identity says: $$(a+5b)^2+(5a-b)^2=26(a^2+b^2)$$

Answer 2

You are looking for $a^2$ and $b^2$, so just create them!
Basically, you know that there are some $\alpha$ and $\beta$ from $\Bbb Z$ such that $a+5b=2002\alpha$ and $5a-b=2002\beta$.
Square your two equations so as to make $a^2$ and $b^2$ appear, then try some things, like adding them...