<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] First, none of them can be in $]-\infty,-2[$ or in $]2,+\infty[$, otherwise it would mean $a^4+b^4+c^4+d^4 > 16$
If some number (let us say $a$) is in $]-2,0[\cup]0,2[$. Then, since $a^4+b^4+c^4+d^4 = 16$ , we have $|b|,|c|,|d|<2$ and $m=\max \{|a|,|b|,|c|,|d|\} < 2$. However, since $a^5+b^5+c^5+d^5\leq m(a^4+b^4+c^4+d^4) < 32$, this is impossible.
So all of them are in $\{2\} \cup\{-2\}$, and $m = 2$.
None of these numbers can be $-2$, because it would mean the other ones are null with $a^4+b^4+c^4+d^4 = 16$.
So at least one of these numbers is $2$. The rest follows.
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