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

As the prime numbers that are $1\bmod4$ can be divided into the sum of two squares, will the squared numbers be unique?

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

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    http://mathworld.wol$f$ram.com/Sumo$f$SquaresFunction.html2012-07-10

3 Answers 3

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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, 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 $n=\sum_{j=1}^{g(k)} a_j^k$](http://math.stackexchange.com/q/103432/19341)2012-07-11
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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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Yes, if you don't take into account the order of the two numbers or $\pm$ sign in front of the numbers.