$a^5+b^5+c^5+d^5=32$ 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?
 A: Suppose $a,b,c,d\geq 0$ and set $x=a^4,\,y=b^4,\,z=c^4,\,w=d^4$. Then
$$f(x,y,z,w)=x^{5/4}+y^{5/4}+z^{5/4}+w^{5/4}$$
is strictly convex and thus restricted to the domain
$$\{x,y,z,w\geq 0: x+y+z+w=16\}$$
it attains the maximum only in the extremal points, which are the four vertices $(16,0,0,0)$ and cyclicals.
If one number is less then zero than taking the moduli the value of $a^5+b^5+c^5+d^5$ strictly increases.
A: 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.
A: We have
$$a^4\le a^4+b^4+c^4+d^4=16\implies a\le 2$$
So, we can have $a^4(a-2)\le 0$, i.e.
$$a^5\le 2a^4$$
Similarly, 
$$b^5\le 2b^4,\quad c^5\le 2c^4,\quad d^5\le 2d^4$$
giving
$$a^5+b^5+c^5+d^5\le 2(a^4+b^4+c^4+d^4)=32$$
Note here that 
$$a^5+b^5+c^5+d^5=2(a^4+b^4+c^4+d^4)$$
holds when
$$a^4(a-2)=b^4(b-2)=c^4(c-2)=d^4(d-2)=0$$
A: 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.
