Elementary Number Theory: Proof

Question

If $\gcd(m,15)= \gcd(n,15)=1$, show that either $15|(m^4 + n^4)$ or $15|(m^4 - n^4)$.

I'm really stuck on this proof. This is what I know:

Since the $\gcd(m,15)= 1$ we can write it as $mx+15y=1$ where $x,y$ are integers. Also, since $\gcd(n,15)=1$ it can also be written as $nu + 15v = 1$ where $u,v$ are integers. I think I'm missing something but I just can't see it.

Answer 1

Hint

As $\phi(15)=8,a^8\equiv1\pmod{15}$ for $(15,a)=1$

$\implies15$ divides $a^8-b^8=?$ for $(15,a)=(15,b)=1$

We can prove something more:$15$ will actually divide $a^4-b^4$

as $\lambda(15)=4, a^4\equiv1\pmod{15}$ for $(a,15)=1$

Answer 2

The conditions mean that $\;m,\,n\;$ aren't divisible by $\;3\;$ and neither by $\;5\;$ , yet $\;m^4\,,\,n^4\;$ are both $\;=1\pmod 3\;$ , whereas by Fermat's Little Theorem both $\;m^4,\,n^4=1\pmod 5\;$ ...