If you want to understand something in my solution then just tell .The part where I have not explained is especially the part where it gets tricky so it may not be very easy but anyway ..... EDIT:I have not posted that part because I am very slow at typing .

I think your solution is wrong kk108 ....

Why ? By the way it is not .....

But if we just plug in your solution (19;7) in the original equation we get: LHS: 19^3-7^3 = 6173 and RHS: 19*7^3-1 = 6516 so either i calculated wrong or your solution is false. Or am I missing something??

$p(p-1)=(q-1)(q^2+q+1)$ Obvious, that $p>q$ , so $p|q^2+q+1$ $q^2+q+1=kp$ $p-1=(q-1)k \to p=kq-k+1$ $q^2+q+1=k^2q-k^2+k$ $q^2+q(1-k^2)+k^2-k+1=0$ $D=k^4-6k^2+4k-3$ Easy to check $(k^2-4)^2<D<(k^2-2)^2 \to D=(k^2-3)^2 \to k=3$ $q^2-8q+7=0 \to q=7 \to p=19$

smashade wrote: But if we just plug in your solution (19;7) in the original equation we get: LHS: 19^3-7^3 = 6173 and RHS: 19*7^3-1 = 6516 so either i calculated wrong or your solution is false. Or am I missing something?? You miscalculated ......

I am very sorry

It's fine ..no worries !

RagvaloD wrote: $p(p-1)=(q-1)(q^2+q+1)$ Obvious, that $p>q$ , so $p|q^2+q+1$ $q^2+q+1=kp$ $p-1=(q-1)k \to p=kq-k+1$ $q^2+q+1=k^2q-k^2+k$ $q^2+q(1-k^2)+k^2-k+1=0$ $D=k^4-6k^2+4k-3$ Easy to check $(k^2-4)^2<D<(k^2-2)^2 \to D=(k^2-3)^2 \to k=3$ $q^2-8q+7=0 \to q=7 \to p=19$ I was so close !!!missed it with a few steps .I missed the last 2s teps .

As it turns out, only the fact that $p$ is prime is needed. Rewriting the equation as $p^3+1=pq^3+q^3\implies (p+1)(p^2-p+1)=q^3(p+1)\implies p^2-p+1=q^3$ and invoking Balkan MO 2005, we're done.