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# Understand the problem

Suppose that for a prime number $p$ and integers $a,b,c$ the following holds:
$$6\mid p+1,\quad p\mid a+b+c,\quad p\mid a^4+b^4+c^4.$$
Prove that $p\mid a,b,c$.

Baltic Way 2009

Number Theory
Medium
##### Suggested Book
Elementary Number Theory by David Burton

Do you really need a hint? Try it first!

Show that, $a^4+b^4+c^4\equiv 2(b^2+bc+c^2)^2$ modulo $p$. Hence $p| (b^2+bc+c^2)$. From there, try to find a relation between $b$ and $c$ modulo $p$.
As $b^2+bc+c^2|b^3-c^3$, we have $b^3\equiv c^3\;\text{mod}\;p$. Note that, $\frac{p+1}{3}$ is an integer and hence $b^{p+1}\equiv c^{p+1}$. Using Fermat’s little theorem, conclude that $b^2\equiv c^2$. Use this along with $b^3\equiv c^3$ to conclude that either $b\equiv c$ or $b,c$ are both divisible by $p$.
Clearly, if $p|b,c$ then we are done.  Prove that the case $b\equiv c$ can be reduced to this one.
If $b\equiv c$, then we have $a\equiv -2b$ and $a^4\equiv -2b^4$. These together give $p|18b^4$. Using $6|p+1$, conclude that $p|b$. The proof is now concluded using hint 3.

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