# 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 $$15\mid m^4-n^4$$.

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 $\gcd(3,5) = 1$, $15 \mid m^4 - n^4$.

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 $3\mid m^4-n^4$

• So you're saying that because it's divisible by 3 and 5 then it must be divisible by 15 right? – Luis Robles Mar 24 '18 at 20:29
• Yes............. – Aqua Mar 24 '18 at 20:30

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.

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?

• No I didn't check 4th powers and that makes sense. Thank you for the advice. – Luis Robles Mar 24 '18 at 20:40

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 $$m^4\equiv (\pm 2^j)^4\equiv (2^4)^j\equiv 1^j \equiv 1 \mod 15 .$$