Do you really need a hint? Try it first!
This Diophantine Equation may seem a bit difficult to handle and will force you to try various techniques like making modulo 7, modulo p, modulo q as p and q are given as primes. But, let’s go through the basic techniques for handling it. So what is it? Checking the parity of p and q in the given equation.
Som check that if both p and q are odd primes, then the LHS will be even but the RHS will be odd, which is a contradiction. Hence the only way it can happen that one of them must be even i.e. 2.
Now, things seem to be under control. We have two cases, p = 2 and q = 2. For p = 2, we get the equation \( q^3 – 14q^2 = -343 \). This implies that q must divide \( 343 = 7^3\). Hence q can be only 7. This gives rise to the solution (2,7). The next hint offers the other case.
For q = 2. we get \(43p^3 – 29p + 9 = 0\). How to solve this? Clearly apply the same idea. Observe that if p = odd the LHS will be odd which can’t be 0. Hence, p must be 2, but it doesn’t satisfy the equation. The only solution is (2,7).