Given that p and q are prime, prove that px2 qx + q = 0 has no rational solutions.
From px2 qx + q = 0 x 0.
Suppose that x = m/n, where HCF(m,n) = 1.
Therefore, pm2/n2 qm/n + q = 0, so pm2 qmn + qn2 = 0.
Dividing through by m, we get pm qn + qn2/m = 0. Clearly m|qn2, as all the other terms are integer. But HCF(m,n) = 1, so m|q. But because q is prime, we deduce that m = 1, q; in the same way n = 1, p.
In which case, x = m/n = 1, 1/p, q, q/p.
If x = 1: p q + q = p 0.
If x = 1/p: 1/p q/p + q = 0. Multiplying by p/q gives 1/q q + p = 0, which is impossible, as all but the first term are integer.
If x = q: pq2 q2 + q = 0. Dividing by q2 gives p 1 + 1/q = 0, which is impossible.
If x = q/p: pq2 p + q = 0. Dividing by q gives pq p/q + 1 = 0, which is impossible.
Hence no rational solutions to the quadratic equation px2 qx + q = 0 exist.
What about the equation, px2 qx + p = 0?