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

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 $144+210 = 354$.
