mathschallenge.net

Problem

Given that p and q are prime, prove that px² qx + q = 0 has no rational solutions.

Solution

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?