<s>[INST] $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero. 
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.

Why does this hold?
 [/INST] Note that (by dividing the equations by $16$ and $32$) this is equivalent to showing one of $a,b,c,d$ is $1$ and the others are $0$ if $a^4 + b^4 + c^4 + d^4 = 1$ and $a^5 + b^5 + c^5 + d^5 = 1$. If none of $a,b,c,d$ are $1$ then they must all have absolute value less than one (otherwise the sum of their fourth powers would exceed $1$). It then suffices to note that
$a^5 + b^5 + c^5 + d^5 < a^4 + b^4 + c^4 +d^4 = 1,$
since $a^5 < a^4, b^5 < b^4, c^5<c^4, d^5<d^4$.
Therefore $a^5 + b^5 + c^5 + d^5 \neq 1$ and we can deduce the result.
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