# 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:

 $\ds 25$ $=$ $\ds 4^2 + \left({2 \times 2^2}\right) + 1^2$ $4$ 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

$\blacksquare$