if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) i want to show that
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) but im not quite sure how to aproach this problem.

Mod $5$, we have that the only squares are $0,1,4$.  The squares of these are $0,1$.  So, we have each of $a^4,b^4,c^4,d^4\in\lbrace 0,1\rbrace $, and their sum is $0$ (becaue $5$ divides them).  Because of this, each of $a^4,b^4,c^4,d^4\equiv 0\pmod{5}$.  Because of this, we have that $a,b,c,d\equiv 0\pmod{5}$, so their sum is divisible by $5$.
An elaboration on "the only squares" argument:
The only numbers mod $5$ are $0,1,2,3,4$.
Lets see how these look like when we square them, and then square them again (take their $4$th power).
\begin{array}{|c|c|c|c|}
\hline
\text{Number}& \text{Square} & \text{Square mod 5} & \text{4th Power} & \text{4th Power mod 5} \\ \hline
0 &0 &0 &0 & 0\\ \hline
1 &1  &1 &1 & 1\\ \hline
2 &4  &4 &16 & 1\\ \hline
3 & 9 & 4 & 16 & 1  \\\hline
4 & 16 & 1 & 1 & 1\\\hline
\end{array}
Look at this last column.  The only possibilities if we take the $4$th power of a number $\pmod{5}$ are $0$ or $1$.
So, each of $a^4,b^4,c^4,d^4$ must be either $0$ or $1$ mod $5$, so if we know they all add up to $0$ (mod $5$), then we know that none of them can be $1$.  The have to all be $0$.
So, we have that $a^4 \equiv b^4\equiv c^4\equiv d^4\equiv 0\pmod{5}$.
So, we know that $a\equiv b\equiv c\equiv d\equiv 0\pmod{5}$ (If we start at the right hand side of the table and move to the left, we see that if something to the $4$th power is $0$ mod $5$, the original thing had to be $0$ mod $5$).  So, we get that $a+b+c+d\equiv 0+0+0+0\equiv 0\pmod{5}$, so it has to be divisible by $5$.
