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Using modular arithmetic, how can one quickly find the natural number n for which $n^5 = 27^5 + 84^5 + 110^5 + 133^5$?

I tried factoring individual components out, but it seemed really tedious.

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    Since when does math have to have a purpose?2012-05-04

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If there is such an $n$, it must be a multiple of 6 and 1 less than a multiple of 5, and it must exceed 133 but not by a whole lot, so my money's on 144.

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    @DavidFaux The calculation shows that, mod $\rm 13,\:$ $\rm\:n^5\equiv 1.\:$ Thus $\rm\:n\not\equiv 0,\:$ so by little Fermat, $\rm\:n^{12}\equiv 1\pmod{13}.\:$ Since $\rm (5,12) = 1\:$ there are integers $\rm\: J,K\:$ such that $\rm\:5J+12K = 1.\:$ Thus $\rm\ n^1\equiv n^{5J+12K}\equiv (n^5)^J (n^{12})^K\equiv 1^J 1^K \equiv 1\:\ (mod\ 13)$ More generally, see my posts on [order ideals.](http://math.stackexchange.com/search?q=user%3A242+%22order+ideal%22&submit=search)2012-05-04
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Tabulating the expression with respect to low primes:

$\bmod 2: 27^5 + 84^5 + 110^5 + 133^5 \equiv 1^5 + 0^5 + 0^5 + 1^5 \equiv 0 \implies n\equiv 0$

$\bmod 3: 27^5 + 84^5 + 110^5 + 133^5 \equiv 0^5 + 0^5 + -1^5 + 1^5 \equiv 0 \implies n\equiv 0$

$\bmod 5: 27^5 + 84^5 + 110^5 + 133^5 \equiv 2^5 + (-1)^5 + 0^5 + (-2)^5 \equiv -1 \implies n^5 \equiv n\equiv -1$

Collecting these gives $n\equiv 24\bmod 30$, which already points at $144$

$\bmod 7: 27^5 + 84^5 + 110^5 + 133^5 \equiv (-1)^5 + 0^5 + (-2)^5 + 0^5 \equiv -1+3 \equiv 2 \equiv n^5\equiv n^{-1} \implies n \equiv 4$

This confirms that $144$ is the only possible solution in range; the next modular equivalence solution would be far out of range at $144+210 = 354$.