2601 as Sum of Three Squares in Ten Different Ways

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Theorem

$2601$ can be expressed as the sum of $3$ squares in $10$ different ways.


Proof

\(\displaystyle 2601\) \(=\) \(\displaystyle 1^2 + 10^2 + 50^2\)
\(\displaystyle \) \(=\) \(\displaystyle 2^2 + 14^2 + 49^2\)
\(\displaystyle \) \(=\) \(\displaystyle 10^2 + 10^2 + 49^2\)
\(\displaystyle \) \(=\) \(\displaystyle 14^2 + 14^2 + 47^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1^2 + 22^2 + 46^2\)
\(\displaystyle \) \(=\) \(\displaystyle 14^2 + 17^2 + 46^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1^2 + 34^2 + 38^2\)
\(\displaystyle \) \(=\) \(\displaystyle 24^2 + 27^2 + 36^2\)
\(\displaystyle \) \(=\) \(\displaystyle 17^2 + 34^2 + 34^2\)
\(\displaystyle \) \(=\) \(\displaystyle 22^2 + 31^2 + 34^2\)

That there are no more can be determined by exhaustion.

$\blacksquare$


Historical Note

In his Curious and Interesting Numbers, 2nd ed. of $1997$, David Wells reports this result as coming from Volume $22$ of Journal of Recreational Mathematics, page $75$, but this has not been corroborated.


Sources