Hello

I've been looking at a question for some time, but haven't been able to find a way to start. The question is:

'If x^2 - px - q = 0, where p and q are positive integers, which of the following could not equal x^3?'

4x + 3, 8x + 5, 8x + 7, 10x + 3, 26x + 5

I know the equation has one positive and one negative root. I've completed the square and found the discriminant, but I can't find a way into the question. Could someone please give me a pointer.

Thank you.

I have a solution, but I'm not sure it's the neatest approach.

$\displaystyle x^2=px+q$

$\displaystyle x^3=px^2+qx$

But as $\displaystyle x^2=px+q$,

$\displaystyle x^3=p(px+q)+qx$

$\displaystyle x^3=p^2x+qx+pq$

$\displaystyle x^3=x(p^2+q)+pq$

We have these options:
$\displaystyle 4x + 3$, $\displaystyle 8x + 5$, $\displaystyle 8x + 7$, $\displaystyle 10x + 3$, $\displaystyle 26x + 5$

First, I checked $\displaystyle 26x+5$

p and q are integers and $\displaystyle pq=5$
This means $\displaystyle p$ and $\displaystyle q$ are $\displaystyle 1$ and $\displaystyle 5$, although I don't know which letter corresponds to which number.

Can I get $\displaystyle p^2+q=26$?
Yes! If $\displaystyle p=5$ and $\displaystyle q=1$! So I can rule this out. Check the others: I found one which didn't work.