## Impossible Quadratic

#### 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?