Diophantine equation of prime numbers Determine all pairs of prime numbers $(p,q)$ that satisfy the equation
$p^3-q^3=pq^3-1$.
I can easily understand that
$p>q$, $q^3=p^2-p+1$, $q^3\equiv 1 \pmod{p}$, $p^3\equiv -1 \pmod{q}$.
But I don't know how to go on. (I know that p=19 and q=7 works)

$p^2 - p + 1 = q^3$
$p(p-1) = (q-1)(q^2+q+1)$.
As you noticed, $p>q$, so $q-1$ is coprime with $p$. So $q-1|p-1$, or instead we can write
$k(q-1)=p-1$ and $kp=q^2 + q + 1$ for some integer $k>1$. 
Get $p$ from the first equation and plug it to the second one:
$k(kq - k + 1) = q^2 + q + 1$
$q^2 + (1-k^2)q +1-k+k^2=0$
So $q$ can only be an integer if discriminant of above equation is a square:
$\Delta = k^4 - 6k^2 + 4k - 3$
But $(k^2 - 3)^2 < \Delta < (k^2 - 2)^2$ for $k>3$.
So we only need to consider $k=2, 3$. For $k=2$, $\Delta$ is not a square. For $k=3$, $\Delta=36$, the quadratic equation has solutions $q=1, 7$. $q=7$ gives $p=19$.
So the only solution for $p, q$ primes is $(19, 7)$.
