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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2http://mathworld.wolfram.com/SumofSquaresFunction.html – anon Jul 10 '12 at 12:17
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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.
M Turgeon
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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.
Glorfindel
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draks ...
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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