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)

Question

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.

Answer 1

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 $5$.

Answer 2

Hint $$a^4-1=(a^2-1)(a^2+1)=(a-1)(a+1)(a^2-4+5)=(a-2)(a-1)(a+1)(a+2)+5(a^2-1)$$

Now, if $5 \nmid a$ one of $(a-2),(a-1),(a+1),(a+2)$ is divisible by $5$.

Answer 3

Remember Eulers theorem applied to the prime $5$ : $$ a^4 = 1 \mod 5 \iff a \mod 5 \neq 0$$ so we have $$(a^4+b^4+c^4+d^4) \mod 5 = 0 \iff a=b=c=d=0 \mod 5$$