If $\gcd(m,15)=\gcd(n,15)=1,$ show that $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$. If $\gcd(m,15)=\gcd(n,15)=1$ show $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$. 
This is what I have so far but I'm stuck on essentially the last step. 
Proof: Assume $\gcd(m,15)=\gcd(n,15)=1$. By the Euler's Phi Function: If $\gcd(a,m)=1$ then $a^{\phi (m)}\equiv 1 \pmod m$. 
$\phi(15)=\phi(5)\phi(3)=(5-1)(3-1)=8$
Hence, $m^8\equiv 1 \pmod{15}$ and $n^8\equiv 1 \pmod{15}$. 
Then, $m^8\equiv 1 \pmod{15}\iff 15\mid m^8-1$. Similarly $15\mid n^8-1$. 
Divisibility property sates if $a\mid b$ and $a\mid c$ then $a\mid b\pm c$.
So $15\mid (m^8-1)-(n^8-1) \Rightarrow 15\mid m^8-n^8$
So I believe I'm right up to this point, please correct if I'm not. I know that $m^8-n^8$ factors to $(m^4+n^4)(m^4-n^4)$ however since $15$ is not a prime number I can't assume $15\mid m^4+n^4$ or $15\mid m^4-n^4$. 
Please help on these last few steps. Thanks. 

Since $\gcd(m,5) = \gcd(n,5)=1$ we have by Fermat little theorem:
$$ m^4\equiv 1\equiv n^4 \pmod 5$$ so $5\mid m^4-n^4$.
On the other hand  since $\gcd(m,3) = \gcd(n,3)=1$ we have again by Fermat little theorem:
$$ m^2\equiv 1\equiv n^2 \pmod 3$$ so $3\mid m^2-n^2$ and so $3\mid m^4-n^4$ 
