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Let $p=4k+3$ be a prime number and $a,b\in\mathbb{N}$ so that $p\mid a^2+b^2$

Prove that $p\mid a$ and $p\mid b$

I tried to use Fermat's little theorem, but I don't know what to do next.

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Math Buster
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    If $a,b\in\bf R$, the division is always possible, whatever $p$! – Bernard Jul 25 '18 at 21:47
  • If $a,b \in \mathbb R$ which is a field, divisibility is immediate - any non-zero element of a field is a unit. – Mark Bennet Jul 25 '18 at 21:47
  • Please show us your work using Fermat's little theorem. We can't help you best without knowing whether you used it correctly, or not. So please edit your question to include this lacking context. – amWhy Jul 25 '18 at 21:55
  • You know Fermat's little theorem. You also know the form of the prime $p$, namely $p=4k+3$, and $p\mid a^2+b^2$. This is where you start. (You should be clear as to why $a$ and $b$ are integers also, viz. previous comments.) – Daniel Buck Jul 25 '18 at 22:05

1 Answers1

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Suppose $a^2 \equiv -b^2 \pmod{p}.$ If $a$ or $b$ is divisible by $p$ then it immediately implies both are divisible by $p.$ So suppose they are not. Then $(\frac{a}{b})^2 \equiv -1 \pmod{p}.$ Which implies $-1$ is a square mod $p.$ However, $(\frac{-1}{p}) = (-1)^{\frac{p-1}{2}}.$ Thus, since $p \equiv 3 \pmod{4}$ this is a contradiction. Hence, both are divisible by $p.$

Another solution: Since $p \equiv 3 \pmod{4}, p$ is a Gaussian prime. Hence $p \mid a^2 + b^2$ implies $p \mid a + bi$ or $p \mid a - bi.$ Thus, $p \mid 2a$ which implies $p \mid a.$ Similarly, $p \mid b.$

Oscar Lanzi
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green frog
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  • I don't understand the part: $(\frac{-1}{p}) = (-1)^{\frac{p-1}{2}}$ and the contradiction. – Math Buster Jul 26 '18 at 00:11
  • $(\frac{a}{p})$ is the Legendre symbol. Look it up. It turns out $(\frac{a}{p}) = a^{\frac{p-1}{2}}.$ As a result, $p \equiv 1 \pmod{4}$ implies $\frac{p-1}{2}$ is even hence $(\frac{-1}{p}) = 1.$ $p \equiv 3 \pmod{4}$ gives you that $\frac{p-1}{2}$ is odd. – green frog Jul 26 '18 at 01:21