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:

\(\ds 25\)

\(=\)

\(\ds 4^2 + \paren {2 \times 2^2} + 1^2\)

$4$ squares

\(\ds \)

\(=\)

\(\ds 3^2 + \paren {4 \times 2^2}\)

$5$ squares

\(\ds \)

\(=\)

\(\ds \paren {2 \times 3^2} + 2^2 + \paren {3 \times 1^2}\)

$6$ squares

\(\ds \)

\(=\)

\(\ds 4^2 + 2^2 + \paren {5 \times 1^2}\)

$7$ squares

\(\ds \)

\(=\)

\(\ds 3^2 + \paren {3 \times 2^2} + \paren {4 \times 1^2}\)

$8$ squares

\(\ds \)

\(=\)

\(\ds \paren {2 \times 3^2} + \paren {7 \times 1^2}\)

$9$ squares

\(\ds \)

\(=\)

\(\ds 4^2 + \paren {9 \times 1^2}\)

$10$ squares

\(\ds \)

\(=\)

\(\ds 3^2 + \paren {2 \times 2^2} + \paren {8 \times 1^2}\)

$11$ 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