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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