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. 
 A: 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 $3\mid m^4-n^4$ 
A: Since none of the existing answers is built-up on OP's work, I posted another one.
OP has proved $15\mid m^8-n^8 = (m^4-n^4)(m^4+n^4)$.
Since $15$ is a composite number, to apply Euclid's Lemma, we have to do so in two steps (on prime number $3$ and $5$).
Note that (without Fermat's Little Theorem)


*

*$a^2 \equiv 0, 1 \pmod 3$, so $a^4 \equiv 0,1 \pmod 3$.


*

*If $3 \mid m^4 + n^4$, it's easy to see that the only possibility is that $3 \mid m,n$, which contradicts $\gcd(m,15) = \gcd(n,15) = 1$

*Use Euclid's Lemma to show that $3 \mid m^4 - n^4$


*$a^2 \equiv 0, 1, 4 \pmod 5$, so $a^4 \equiv 0,1 \pmod 5$, and $m^4 + n^4 \equiv 0,1,2 \pmod 5$.


*

*If $5 \mid m^4 + n^4$, its again easy to see that the only possibility is $5 \mid m,n$, which contradicts $\gcd(m,15) = \gcd(n,15) = 1$

*Use Euclid's Lemma to show that $5 \mid m^4 - n^4$



Since $\gcd(3,5) = 1$, $15 \mid m^4 - n^4$.
A: This question looks familiar... At any rate, the conclusion $15 \mid (m^4 + n^4)$ is unnecessary. As noted before, applying Fermat's Little Theorem with the primes 3 and 5 is all you need to solve this question.
A: Good work! Have you checked the $4$th powers $\mod 15$? For what numbers (with $\gcd(k,15)=1$) can we say $k^4\equiv 1\mod 15$? Can you use that to finish your proof?
A: If $m$ is co-prime to $15$ then $m$ is congruent modulo $15$ to a member of $\{\pm 1, \pm 2, \pm 4,\pm 8\}$ so for some $j\in \{0,1,2,3\}$ we have $$m^4\equiv (\pm 2^j)^4\equiv (2^4)^j\equiv 1^j \equiv 1 \mod 15 .$$
