If $a$ and $c$ are odd primes and $ax^2+bx+c=0$ has rational roots, where $b\in I$, then prove that one root will be independent of $a$ and $b$.

Question

I am having trouble with this tricky question on number theory and quadratic equations.

If $a$ and $c$ are odd prime numbers and $ax^2+bx+c=0$ has rational roots, where $b\in I$, then prove that one root of the equation will be independent of $a$ and $b$.

So I thought of trying the following way:

If it has rational roots (say $u$ and $m$) then $u . m = a . c$ and u + m = b. As $a$ and $c$ are prime, $a + c = b$, and plugging this into the quadratic formula, we get one root will always be $-1$.

I am not sure if this proof is correct or if my final value is correct. Could someone please give a detailed proof?

P.S.: I couldn’t find any examples regarding my question so if you do, please include them too.

Thank You.

Answer 1

I think you are almost correct (assuming you mean by $I$ the set of integers). The only thing you miss is that if $a,c$ are prime, then you have $a+c=\pm b$ (which follows from the fact that $b^2-4ac$ must be the square of a rational number), so the possible values are $\pm 1$. You can get as many examples as you want just from the formula $a+c=\pm b$ you have (e.g. $(a,b,c)=(3,\pm 8,5)$).

P.s. Your formulas u * m = a * c and u + m = b look wrong, but if you have $a+c=\pm b$, your quadratic equation is just $ax^2 \pm (a+c) x + c =0$, which you can just solve and is essentially the same as using the formula you want to apply.

Answer 2

Since $ \ a \ $ and $ \ c \ $ are odd prime numbers, the Rational Zeroes Theorem proposes the candidates $$ \ x \ = \ \pm \frac{c}{a} \ , \ \pm \frac{c}{1} \ , \ \pm \frac{1}{a} \ , \ \pm \frac{1}{1} \ \ . $$

The factorization of the quadratic polynomial is $ \ ax^2 \ + \ bx \ + \ c \ = \ a·(x - r)·(x - s) \ \ , $ so the four factorizations that will produce the correct leading and constant terms are $$ a·\left(x - \frac{c}{a} \right)·\left(x - 1 \right) \ \ , \ \ a·\left(x + \frac{c}{a} \right)·\left(x + 1 \right) \ \ , \ \ a·\left(x - \frac{1}{a} \right)·\left(x - c \right) \ \ , $$ $$ a·\left(x + \frac{1}{a} \right)·\left(x + c \right) \ \ . $$

The "middle coefficient" $ \ b \ = \ -a·(r + s) \ $ in these products of factors is one of $$ -a · \left(\pm \frac{c}{a} \ \pm \ 1 \right) \ \ = \ \ \pm \ (c + a) \ \ \ \text{or} \ \ \ -a · \left(\pm \frac{1}{a} \ \pm \ c \right) \ \ = \ \ \pm \ (1 + ac) \ \ , $$ any of which is an integer.

Hence, one of the rational roots of the quadratic equation is $ \ \pm 1 \ $ or $ \ \pm c \ \ , $ which are independent of $ \ a \ $ and $ \ b \ \ . $