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 $p\mid a,b,c$.

Source of the problem

Baltic Way 2009

Topic
Number Theory
Difficulty Level
Medium
Suggested Book
Elementary Number Theory by David Burton

Start with hints

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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