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:
\(\ds 25\) | \(=\) | \(\ds 4^2 + \left({2 \times 2^2}\right) + 1^2\) | $4$ squares | |||||||||||
\(\ds \) | \(=\) | \(\ds 3^2 + \left({4 \times 2^2}\right)\) | $5$ squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({2 \times 3^2}\right) + 2^2 + \left({3 \times 1^2}\right)\) | $6$ squares | |||||||||||
\(\ds \) | \(=\) | \(\ds 4^2 + 2^2 + \left({5 \times 1^2}\right)\) | $7$ squares | |||||||||||
\(\ds \) | \(=\) | \(\ds 3^2 + \left({3 \times 2^2}\right) + \left({4 \times 1^2}\right)\) | $8$ squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({2 \times 3^2}\right) + \left({7 \times 1^2}\right)\) | $9$ squares | |||||||||||
\(\ds \) | \(=\) | \(\ds 4^2 + \left({9 \times 1^2}\right)\) | $10$ squares | |||||||||||
\(\ds \) | \(=\) | \(\ds 3^2 + \left({2 \times 2^2}\right) + \left({8 \times 1^2}\right)\) | $11$ squares |
$\blacksquare$
Sources
- Feb. 1993: Kelly Jackson, Francis Masat and Robert Mitchell: Extensions of a Sums-of-Squares Problem (Math. Mag. Vol. 66, no. 1: pp. 41 – 43) www.jstor.org/stable/2690474