Numbers Partitioned into up to 4 Squares in 5 Ways

From ProofWiki
Jump to navigation Jump to search

Theorem

The following positive integers can be expressed as the sum of no more than $4$ squares in $5$ distinct ways:

$50, 52, 54, 58, \ldots$


Proof

\(\ds 50\) \(=\) \(\ds 7^2 + 1^2\)
\(\ds \) \(=\) \(\ds 5^2 + 5^2\)
\(\ds \) \(=\) \(\ds 5^2 + 4^2 + 3^2\)
\(\ds \) \(=\) \(\ds 4^2 + 4^2 + 3^2 + 3^2\)
\(\ds \) \(=\) \(\ds 6^2 + 3^2 + 2^2 + 1^2\)


\(\ds 52\) \(=\) \(\ds 7^2 + 1^2 + 1^2 + 1^2\)
\(\ds \) \(=\) \(\ds 6^2 + 4^2\)
\(\ds \) \(=\) \(\ds 5^2 + 5^2 + 1^2 + 1^2\)
\(\ds \) \(=\) \(\ds 4^2 + 4^2 + 4^2 + 2^2\)
\(\ds \) \(=\) \(\ds 5^2 + 3^2 + 3^2 + 3^2\)


\(\ds 54\) \(=\) \(\ds 7^2 + 2^2 + 1^2\)
\(\ds \) \(=\) \(\ds 6^2 + 4^2 + 1^2 + 1^2\)
\(\ds \) \(=\) \(\ds 6^2 + 3^2 + 3^2\)
\(\ds \) \(=\) \(\ds 5^2 + 5^2 + 2^2\)
\(\ds \) \(=\) \(\ds 5^2 + 4^2 + 3^2 + 2^2\)


\(\ds 58\) \(=\) \(\ds 7^2 + 3^2\)
\(\ds \) \(=\) \(\ds 7^2 + 2^2 + 2^2 + 1^2\)
\(\ds \) \(=\) \(\ds 6^2 + 3^2 + 3^2 + 2^2\)
\(\ds \) \(=\) \(\ds 5^2 + 5^2 + 2^2 + 2^2\)
\(\ds \) \(=\) \(\ds 5^2 + 4^2 + 4^2 + 1^2\)

$\blacksquare$


Sources