MHB Prove a sum is not the fifth power of any integer

[anemone]

Suppose $X$ is a number of the form $\displaystyle X=\sum_{k=1}^{60} \epsilon_k \cdot k^{k^k}$, where each $\epsilon_k$ is either 1 or -1.

Prove that $X$ is not the fifth power of any integer.

 

[anemone]

Solution of other:

Spoiler

WLOG, we may assume that $\epsilon_{60}=1$. Let $\large m=60^{\dfrac{1}{5}({60^{60})}}$.

We show that if $\epsilon_{59}=-1$, then $(m-1)^5<X<m^5$---(1) and

if $\epsilon_{59}=1$, then $m^5<X<(m+1)^5$---(2)

From (1), we note that $60^{60}=(59+1)^{60}>59^{60}+60(59)^{59}>2\cdot 59^{60}$---(3)

so that using (3)

$\large m^3=60^{\dfrac{3}{5}({60^{60})}}>60^{\dfrac{6}{5}({59^{60})}}>59^{({59^{60})}}>59^{({59^{59+1})}}=59\cdot59^{59^{59}}$----(4)

Then

$\begin{align*}(m-1)^5&=m^5-5m^3(m-1)-5m(2m-1)-1\\&<m^5-5m^3(m-2)\\&<m^5-5m^3\\&<m^5-m^3\end{align*}$

and from (4) we have

$\begin{align*}m^5-m^3&<m^5-59\cdot59^{59^{59}}\\&<60^{60^{60}}+\sum_{k=1}^{59} (-1)k^{k^k}\\&\le X\\&<m^5-59^{59^{59}}+58\cdot 58^{58^{58}}\\&<m^5\end{align*}$

A similar argument proves (2) and we're done.