As 110 and 84 are even, and 27 and 133 are odd, $2|n^5$. Both 84 and 27 are divisible by 3, and $3|(133+110)|133^5+110^5$, so $3|n^5$. By FLT $n^5 \equiv n \mod 5$, and by repeated use and addition, $n^5 \equiv 4 \mod 5$ and so $n \equiv 4 \mod 5$. Both 133 and 84 are divisible by 7, $27^5 \equiv -1 \mod 7$, and $110^5 \equiv -32 \equiv 3 \mod 7$. Hence $n^5 \equiv 2 \mod 7$. We can then get $n \equiv 4 \mod 7$. By CRT, $n \equiv 144 \mod 210$. By some really weak bounds, we can say that $n=144$ specifically.

Moonmathpi496 wrote: n is obviously even.Now check the last digits of rhs(mod 10). Rhs=3+0+4+7==4 so last digit of n will be 4.(bcz 2^5==2) not we set a lower bound.Observe that rhs==0 mod 3. k>133. So possibilities are 144,174. Now 174^5=(133+41)^5.Using binomial theorem,we may show that it is >rhs. So 144 is the answer. Moon

Довольно интересно, что если мы захотим искать решения в целых комплексных числах, то можно даже получить формулу решения уравнения такого вида. И число слагаемых можно уменьшить. Там опечатка сейчас только заметил. Конечно $b=p^2+2ps-s^2$

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Answer is 144 which we get by applying modular arithmetic

Note that $2\mid n$ and $3\mid n,$ also it is clear that $n\equiv 4\pmod 5.$ So $n>133,$ which means $n=144$ or $n=174,$ it is not hard to disregard the latter.