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Suppose that $p$ is a prime of the form $4k + 1$. I'm trying to prove that there exist nonzero integers $a$ and $b$ such that $p^2 = a^2 + b^2$.

I begin by noting that $p$ itself can be written as the sum of two squares, since it does not have any prime factors of the form $4k+3$ (this is a theorem from my discrete math textbook: a positive integer $n$ can be represented as the sum of two squares if and only if every prime divisor of $n$ of the form $4k+3$ appears in the canonical representation of $n$ with an even exponent).

We thus have $p=u^2+v^2$, and $p^2 = u^4+2u^2v^2+v^4$. I now need to prove that the right-hand side is the sum of two squares, but I'm lost. Obviously, it can be written as the sum of four squares, but that's not what I'm trying to show. Any hint or nudge in the right direction would be appreciated.

Barry Powell
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Eragon20
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  • Hint: for any two complex numbers $a,b$, $|a|^2|b|^2=|c|^2$ where $c=ab$. – Aphelli Apr 12 '21 at 20:18
  • Are you trying to prove the statement without using the theorem from your textbook? because that theorem implies it immediately. – Greg Martin Apr 12 '21 at 20:23
  • @GregMartin Just realized how it follows from simple algebra, which is all I was looking for. However, a proof without the theorem would be fascinating in itself. – Eragon20 Apr 12 '21 at 20:28
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    More generally products oif sums of squares are sum of squares, due to the well known composition law (Brahmagupta–Fibonacci), (or equivalently multiplicativity of norms of Gaussian integers (cf. my answer in the dupe). OP is a special case when the product is a square. – Bill Dubuque Apr 12 '21 at 20:46
  • All Pythagorean triples have integer solutions where $a^2+B^2=C^2$ and $C=4n+1, n\in\mathbb{N}$ – poetasis Jul 31 '21 at 23:31

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If $p = u^2 + v^2$, then we can write $p^2 = (u^2 - v^2)^2 + (2uv)^2$.

The idea behind this is to use the norm map on $\Bbb Z[i]$ (or simply on $\Bbb C$). That is, $p = |u + iv|^2$ implies $p^2 = |(u + iv)^2|^2$.

WhatsUp
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Have you heard of the Lagrange identity in two variables? If not, see if you can cleverly use a variant of the Sophie Germain identity. If you are still confused, try thinking about $x^2+2xy+y^2 = (x^2-2xy+y^2)+4xy$. Hmm...

Alien
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  • Thanks for this. I now see how this follows from simple algebraic manipulation, and also appreciate the recourse to the Lagrange identity as a theoretical background. – Eragon20 Apr 12 '21 at 20:26