Fourth Powers which are Sum of 4 Fourth Powers/Examples/353

From ProofWiki
Jump to: navigation, search

Examples of Fourth Powers which are Sum of 4 Fourth Powers

$353^4 = 30^4 + 120^4 + 272^4 + 315^4$


Proof

\(\displaystyle 30^4 + 120^4 + 272^4 + 315^4\) \(=\) \(\displaystyle 810 \, 000\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle 207 \, 360 \, 000\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle 5 \, 473 \, 632 \, 256\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle 9 \, 845 \, 600 \, 625\)
\(\displaystyle \) \(=\) \(\displaystyle 15 \, 527 \, 402 \, 881\)
\(\displaystyle \) \(=\) \(\displaystyle 353^4\)


Now we have that:

\(\displaystyle 442^2 - 272^2\) \(=\) \(\displaystyle 170 \times 714\)
\(\displaystyle \) \(=\) \(\displaystyle 17^2 \times 420\)

Hence:

\(\displaystyle 442^2 - 3 \times 17^2\) \(=\) \(\displaystyle 272^2 + 289 \times 417\)
\(\displaystyle \) \(=\) \(\displaystyle 272^2 + 353^2 - 64^2\)

But:

$3 \times 17 = 2 \times 26 - 1$

So:

\(\displaystyle 442^2 - 2 \times 26 \times 17 + 17\) \(=\) \(\displaystyle 442^2 - 2 \times 442 + 17\)
\(\displaystyle \) \(=\) \(\displaystyle 441^2 + 4^2\)
\(\displaystyle \) \(=\) \(\displaystyle 21^4 + 2^4\)
\(\displaystyle \) \(=\) \(\displaystyle 272^2 + 353^2 - 8^4\)

Hence:

$353^2 + 272^2 = 2^4 + 8^4 + 21^4$

but:

\(\displaystyle 353^2 - 272^2\) \(=\) \(\displaystyle 81 \times 625\)
\(\displaystyle \) \(=\) \(\displaystyle 15^4\)

So:

$353^4 = 30^4 + 120^4 + 272^4 + 315^4$

$\blacksquare$


Historical Note

This result was discovered by Robert Norrie, who reported on it in $1911$.


Sources