Confusion? (Continued)

Number Theory Level 2

Given :

$p$ is an odd prime number.
$x, y$ are positive integers.
$p=x^2+y^2$ .

What can you conclude from this?

p\equiv {-1}\pmod 4

p\equiv 1\pmod 4

Not enough information

1 solution

Chris Lewis
Feb 13, 2020

Since $p$ is odd, $x$ and $y$ can't have the same parity. Without loss of generality, say $x$ is odd and $y$ is even. Then $x^2 \equiv 1 \pmod4$ and $y^2 \equiv 0 \pmod4$ , so that $p \equiv 1 \pmod4$ .

A serious typo! "say $x$ is even and... " should be "say $x$ is odd and $y$ is even ...". The solution is excellent, seems better than Zagier's theorem.

A Former Brilliant Member - 1 year, 3 months ago

Aha - I thought these problems might be heading toward Zagier - carry on!!

Chris Lewis - 1 year, 3 months ago

FWIW, the result only uses the odd parity of p and not the fact that it's a prime number.

Richard Desper - 1 year, 3 months ago

Confusion? (Continued)

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