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There's the Fermat's theorem on sums of two squares.

My question is, as the prime numbers that are 1 modulo 4 can be divided into the sum of two squares, will the squared number be unique?

For example, $41 = 4^2 + 5^2$ and the squared number will be 4 and 5.

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You wrote: as the numbers that are 1 modulo 4 can be divided into the sum of two squares. Did you want to write: as the prime numbers that are...? –  Martin Sleziak Jul 10 '12 at 12:16
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@MartinSleziak right. I forgot to write that :) –  Zat Mack Jul 10 '12 at 12:18

3 Answers 3

up vote 5 down vote accepted

Yes, if you don't take into account the order of the two numbers or $\pm$ sign in front of the numbers.

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Just to complement Pantelis' answer, the reason why they are unique can be easily seen from the proof using the Gaussian integers $\mathbb{Z}[i]$, which is a UFD.

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Primes of the form $p=4k+1\;$ have a unique decomposition as sum of squares $p=a^2+b^2$ with $0<a<b\;$, due to Thue's Lemma.

Further, primes of the form $p=4n+3$, never have a decomposition into $2$ squares, proven in various ways here.

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BTW: Unique solutions seem to be quite rare, as this unanswered question indicates: Unique solutions for $n=\sum_{j=1}^{g(k)} a_j^k$ –  draks ... Jul 11 '12 at 10:41

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