2601 as Sum of 3 Squares in 12 Different Ways

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Theorem

$2601$ can be expressed as the sum of $3$ squares in $12$ different ways.


Proof

\(\displaystyle 2601\) \(=\) \(\displaystyle 51^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1^2 + 10^2 + 50^2\)
\(\displaystyle \) \(=\) \(\displaystyle 2^2 + 14^2 + 49^2\)
\(\displaystyle \) \(=\) \(\displaystyle 10^2 + 10^2 + 49^2\)
\(\displaystyle \) \(=\) \(\displaystyle 14^2 + 14^2 + 47^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1^2 + 22^2 + 46^2\)
\(\displaystyle \) \(=\) \(\displaystyle 14^2 + 17^2 + 46^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1^2 + 34^2 + 38^2\)
\(\displaystyle \) \(=\) \(\displaystyle 14^2 + 31^2 + 38^2\)
\(\displaystyle \) \(=\) \(\displaystyle 3^2 + 36^2 + 36^2\)
\(\displaystyle \) \(=\) \(\displaystyle 24^2 + 27^2 + 36^2\)
\(\displaystyle \) \(=\) \(\displaystyle 17^2 + 34^2 + 34^2\)
\(\displaystyle \) \(=\) \(\displaystyle 22^2 + 31^2 + 34^2\)

That there are no more can be determined by exhaustion.

$\blacksquare$


Sources

  • 1989: John M. HowellProblems and Conjectures: $\text Q 1692$. Three Squares (J. Recr. Math. Vol. 21, no. 1: p. 68)
  • 1990: Solutions to Problems and Conjectures: $\text Q 1692$. Three Squares (J. Recr. Math. Vol. 22, no. 1: pp. 74 – 76)