mathschallenge.net

Problem

Given that $p$ and $q$ are prime, prove that $px$² $qx$ + $q$ = 0 has no rational solutions.

Solution

From $px$² $qx$ + $q$ = 0 $x$ 0.

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

Therefore, $pm$²/$n$² $qm$/$n$ + $q$ = 0, so $pm$² $qmn$ + $qn$² = 0.

Dividing through by $m$, we get $pm$ $qn$ + $qn$²/$m$ = 0. Clearly $m$|$qn$², 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 $x$ = $q$: $pq$² $q$² + $q$ = 0. Dividing by $q$² gives $p$ 1 + 1/$q$ = 0, which is impossible.

If $x$ = $q$/$p$: $pq$² $p$ + $q$ = 0. Dividing by $q$ gives $pq$ $p$/$q$ + 1 = 0, which is impossible.

Hence no rational solutions to the quadratic equation $px$² $qx$ + $q$ = 0 exist.

What about the equation, $px$² $qx$ + $p$ = 0?