Find all primes p and q such that \dfrac{p^3-2017}{q^3-345}=q^3.
Proposed by Titu Andreescu, University of Texas at Dallas, USA
Solution:
The given equation is equivalent to p^3-2017=q^6-345q^3. Reducing modulo 7, we get p^3-1 \equiv q^6-2q^3 \pmod{7} \iff p^3 \equiv (q^3-1)^2 \pmod{7}.
Since q^3 \equiv 0,1,6 \pmod{7}, then (q^3-1)^2 \equiv 0,1,4 \pmod{7}. Since p^3 \equiv 0,1,6 \pmod{7}, we get (q^3-1)^2 \equiv 0,1 \pmod{7}. If (q^3-1)^2 \equiv 0 \pmod{7}, then p^3 \equiv 0 \pmod{7}, which gives p=7. But the equation q^6-345q^3+1674=0 gives no integer solutions. If (q^3-1)^2 \equiv 1 \pmod{7}, then q^3 \equiv 0 \pmod{7}, which gives q=7. So, p^3-2017=-686, i.e. p^3=1331, which gives p=11. So, p=11 and q=7.
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