25 as Sum of 4 to 11 Squares

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Theorem

$25$ can be expressed as the sum of $n$ non-zero squares for all $n$ from $4$ to $11$.


Proof

We have:

\(\displaystyle 25\) \(=\) \(\displaystyle 4^2 + \left({2 \times 2^2}\right) + 1^2\) $4$ squares
\(\displaystyle \) \(=\) \(\displaystyle 3^2 + \left({4 \times 2^2}\right)\) $5$ squares
\(\displaystyle \) \(=\) \(\displaystyle \left({2 \times 3^2}\right) + 2^2 + \left({3 \times 1^2}\right)\) $6$ squares
\(\displaystyle \) \(=\) \(\displaystyle 4^2 + 2^2 + \left({5 \times 1^2}\right)\) $7$ squares
\(\displaystyle \) \(=\) \(\displaystyle 3^2 + \left({3 \times 2^2}\right) + \left({4 \times 1^2}\right)\) $8$ squares
\(\displaystyle \) \(=\) \(\displaystyle \left({2 \times 3^2}\right) + \left({7 \times 1^2}\right)\) $9$ squares
\(\displaystyle \) \(=\) \(\displaystyle 4^2 + \left({9 \times 1^2}\right)\) $10$ squares
\(\displaystyle \) \(=\) \(\displaystyle 3^2 + \left({2 \times 2^2}\right) + \left({8 \times 1^2}\right)\) $11$ squares

$\blacksquare$


Sources