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Impossible Quadratic

Problem

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


Solution

From px2 minus qx + q = 0 implies x greater than 0.

Suppose that x = m/n, where HCF(m,n) = 1.

Therefore, pm2/n2 minus qm/n + q = 0, so pm2 minus qmn + qn2 = 0.

Dividing through by m, we get pm minus 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 minus q + q = p not equal 0.

If x = 1/p: 1/p minus q/p + q = 0. Multiplying by p/q gives 1/q minus q + p = 0, which is impossible, as all but the first term are integer.

If x = q: pq2 minus q2 + q = 0. Dividing by q2 gives p minus 1 + 1/q = 0, which is impossible.

If x = q/p: pq2 minus p + q = 0. Dividing by q gives pq minus p/q + 1 = 0, which is impossible.

Hence no rational solutions to the quadratic equation px2 minus qx + q = 0 exist.

What about the equation, px2 minus qx + p = 0?

Problem ID: 244 (19 Oct 2005)     Difficulty: 4 Star

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