25 as Sum of 4 to 11 Squares

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

• Feb. 1993: Kelly Jackson, Francis Masat and Robert Mitchell: Extensions of a Sums-of-Squares Problem (Math. Mag. Vol. 66, no. 1: 41 – 43)  www.jstor.org/stable/2690474