Numbers Partitioned into up to 4 Squares in 5 Ways
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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
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $50$