One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$.
Taking the given equation modulo $2,3,$ and $5,$ respectively, we have
\begin{align*} n^5&\equiv0\pmod{2}, \\ n^5&\equiv0\pmod{3}, \\ n^5&\equiv4\pmod{5}. \end{align*}
By either Fermat's Little Theorem (FLT) or inspection, we get
\begin{align*} n&\equiv0\pmod{2}, \\ n&\equiv0\pmod{3}, \\ n&\equiv4\pmod{5}. \end{align*}
By either the Chinese Remainder Theorem (CRT) or inspection, we get $n\equiv24\pmod{30}.$
It is clear that $n>133,$ so the possible values for $n$ are $144,174,204,\ldots.$ Note that
\begin{align*} n^5&=133^5+110^5+84^5+27^5 \\ &1.3^5 \\ &=1.3^2\cdot1.3^2\cdot1.3 \\ &>1.6\cdot1.6\cdot1.3 \\ &=2.56\cdot1.3 \\ &>2.5\cdot1.2 \\ &=3, \end{align*}
which arrives at a contradiction. Therefore, we conclude that $n=\boxed{144}.$
~MRENTHUSIASM