hint

$a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$ so obviously if $p|a-b$ then $p|a^3-b^3$ Suppose $p|a^3-b^3$ Obviously if $p|a$ or $p|b$ then $p$ must divide both of them and thus $p|a-b$ so assume that $p$ is not a factor of $a$ or $b$ Then by Fermat's Little theorem $a^{p-1}=b^{p-1}=1\left(\operatorname{mod}p\right)$ We can rewrite this as $a\cdot\left(a^3\right)^{\frac{p-2}{3}}=b\cdot\left(b^3\right)^{\frac{p-2}{3}}\left(\operatorname{mod}p\right)$ and because from our assumption $a^3=b^3\left(\operatorname{mod}p\right)$ is follows that $a=b\left(\operatorname{mod}p\right)$ and thus $p|a-b$

Kaskade wrote: $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$ so obviously if $p|a-b$ then $p|a^3-b^3$ Suppose $p|a^3-b^3$ Obviously if $p|a$ or $p|b$ then $p$ must divide both of them and thus $p|a-b$ so assume that $p$ is not a factor of $a$ or $b$ Then by Fermat's Little theorem $a^{p-1}=b^{p-1}=1\left(\operatorname{mod}p\right)$ We can rewrite this as $a\cdot\left(a^3\right)^{\frac{p-2}{3}}=b\cdot\left(b^3\right)^{\frac{p-2}{3}}\left(\operatorname{mod}p\right)$ and because from our assumption $a^3=b^3\left(\operatorname{mod}p\right)$ is follows that $a=b\left(\operatorname{mod}p\right)$ and thus $p|a-b$ WOW! I would not have thought of THAT manipulation.

the forward direction is obvious. suppose for contradiction that $p | a-b$ and p doesn't divide $a^2+ab+b^2$. If a = 0 mod p then p | b i.e. b = 0 mod p giving a-b = 0 mod p. So b isn't 0 mod p hence b has a multiplicative inverse mod p $$p | a^2+ab+b^2$$multiplying by $b^{-1}^2$ and substituting $z=ab^-1$ $$p | z^2+z+1$$multiplying RHS by $z-1$: $$p | z^3-1$$which means that $ord_p(z)=3$ or z = 1 but if z=1 then we have p | 3 i.e p=3 which isn't true, otherwise we have $ord_p(z) = 3 | p-1$ which contradicts our initial assumption that $3 | p+1$.

RANDOMMATHLOVER wrote: WOW! I would not have thought of THAT manipulation. Well, that is quite an intuitive manipulation. Just to ensure that you feel the same, here is another question that uses the same: Prove that if there is a prime $p$ such that $p|a^2+b^2$ and $4|p+1$, then $p|a$ and $p|b$. $a,b \in \mathbb{N}$ ofcourse. Also @above, nice solution. You're anything but a WeakMathematician @below yes, the techniques of both solutions are applicable here And if you look at them closely, they really are the same thing!

MathPassionForever wrote: RANDOMMATHLOVER wrote: WOW! I would not have thought of THAT manipulation. Well, that is quite an intuitive manipulation. Just to ensure that you feel the same, here is another question that I want the same: Prove that if $p|a^2+b^2$ and $4|p+1$, then $p|a$ and $p|b$. suppose the contrary i.e. that p doesn't divide b. then there is a multiplicative inverse of b modulo p i.e.: $$a^2+b^2 \equiv 0 mod p$$$$a^2 b^{-2} + 1 \equiv 0 mod p$$$$z^2+1 \equiv 0 mod p$$$$z^4 \equiv 1 mod p$$$$\rightarrow 4 | p-1$$a contradiction. So no multiplicative inverse can exist ==> p|a,b

Devastator wrote: Let $p$ be a prime such that $3|p+1$. Show that $p|a-b$ if and only if $p|a^3-b^3$ Overkill maybe? Clearly the "only if" direction is trivial so lets go by the "if" direction. Case 1: $p$ divides $a$ or $b$ If $p$ divides $a$ or $b$ then it divides both so that means $p \mid a-b$ as desired. Case 2: $p$ doesnt divide $a$ or $b$ Now by contradiction assume that $p \not \; \mid a-b$ so $p \mid a^2+ab+b^2$ but that means $p \mid \Phi_3(a,b)$ so either $p=3$ or $3 \mid p-1$ but both cases yeld contradictions, hence $p \mid a-b$ as desired.

Assume to the contrary, $$p|a^3-b^3 \implies p|a^2+ab+b^2 \implies p|(2a+b)^2+3b^2$$So $-3$ is a QR modulo $p$.Thus, $$1=(\frac{-3}{p})=(\frac{-1}{p}).(\frac{3}{p})=(-1)^{\frac{p-1}{2}}.(-1)^{\frac{p-1}{2}.\frac{3-1}{2}}.(\frac{p}{3}) \stackrel{p \equiv -1 \pmod 3}=(-1)^{p-1}.-1=-1$$Which is a clear contradiction and we are done