Consequence following from the hypothesis $ : 7$ divides $a^2 + b^2$ with $ a, b \in \mathbb Z$

Question

Not HM, simply self teaching.

Source : Alain Troesch (Louis-le-Grand High School, Paris) , Exercices $2022-2023$, Polycopié des exercices , page $99$, Ex. $21.10$ http://alain.troesch.free.fr/

Given ( by hypothesis) that $7 | a^2 + b^2$ with $a, b \in \mathbb Z$. Prove that $ 7$ divides both $a$ and $b$.

I only manage to prove the implication $ 7|a \implies 7|b^2$ but not $ 7|a \implies 7|b$ , even less the desired conjunction.

Proof of the implication :

Since $7 | a^2 + b^2$ , it follows that: $ a^2 + b^2 = 7k , k \in \mathbb Z$.

Now suppose that $7|a$ meaning equivalently that : $ a = 7k'$.

In that case we have :

$(7k')^2 + b^2 = 7k$

$\iff 7^2(k')^2 + b^2 = 7k $

$\iff b^2 = 7[ k -7(k')^2]$

$ \iff 7 | b^2 $

Answer 1

One of the properties of primes is that

Suppose a prime $p\mid ab$ then $p\mid a$ or $p\mid b$

This follows from the fundamental theorem of arithmetic.

Now since $7\mid b\cdot b$ then either $7\mid b$ or $7\mid b$. So $7\mid b$.

Notice that you have only proven that if $7\mid a$ then $7\mid b$. The case still remains where $7\not\mid a,b$. You can now write:

• $a=7m+x$, $1\leq x<7$
• $b=7n+y$, $1\leq y<7$

Then after some algebra $$\begin{align}a^2+b^2&=(7m+x)^2+(7n+y)^2\\&=x^2+y^2+7(7m^2+7n^2+2mx+2my)\end{align}$$

Can you take it from here?