I'm stuck on this question. I tried, but don't know how to proceed:
Given that $p$ and $q$ are primes, prove that $p + q = (p − q)^3$ if and only if $p = 5$ and $q = 3$.
Is there a way to solve this equation through a congruence relation?
I start by setting:
$(p − q)^3 \equiv 0 \ \mod \ p + q$
On
$$(p+q)=(p-q)^3$$ let suppose that $p=q=2$. then we have: $$4=(0)^3-0$$ which is impossible.
Let suppose wlog that $p=2$ and $q$ is odd. Then $p-2$ and $p+2$ are co-prime therefore $$(p+2)\neq (p-2)^3$$
Let $p$ and $q$ be 2 odd primes. Since $p$ and $q$ are prime then their sum and difference are even. we set $$p+q=2r$$ $$p-q=2s$$ It follows that $$p=(r+s)$$ $$q=r-s$$ Futhermore $$2r=8s^3$$ $$r=4s^3$$ Since $\gcd(p,q)=1$ then $s=1$ Hence $$r=4$$ It follows $$p=4+1=5$$ $$q=4-1=3$$
Hint $\ p\!-\!q\mid p\!+\!q\,\Rightarrow\, p\!-\!q\mid 2p,2q\,\Rightarrow\, p\!-\!q\mid (2p,2q) = 2(p,q) = 2$