Divisibility exercise

Question

Now, I have an idea how to attempt this question with modulo arithmetic, but I was thinking if there was a solution that did not involve modular arithmetic.

If $7 |(b^2+c^2)$ iff $7|b$ and $7|c$.

I can prove it in the reverse direction $(\Leftarrow )$.

We simply use the fact that if $a|b$ and $a|c$, then we have $a|kb+lc$ for some integers $k$ and $l$.

So let $a = 7, k = b, l = c$ and we have our result.

How do we solve it in the other direction?

Without modular arithmetic

Answer 1

As requested for "Without modular arithmetic"

Any $n=7k,7k\pm1,7k\pm2,7k\pm3$ where $k$ is any integer

$\implies n^2=49k^2,49k^2\pm14k+1,49k^2\pm28k+4,49k^2\pm42k+9$

Observe that if $7\nmid bc, b^2,c^2$ will have to be of the form $\in \{49k^2\pm14k+1,49k^2\pm28k+4,49k^2\pm42k+9\}$

Observe that sum of no two is divisible by $7$

Answer 2

$p\equiv3\pmod 4, p|(a^2+b^2)$, then $p|a,p|b$

in fact, if $p|(a^2+b^2)$, we have $$a^2\equiv-b^2 \pmod p$$

so

$$ (\frac ap)^2=(\frac {-1}p)(\frac bp)^2$$ that is $$ (\frac ap)^2=-(\frac bp)^2$$

we must have

$$ (\frac ap)=(\frac bp)=0$$

$p|a,p|b$