Hello,

I was thinking over the following problem. Let $\displaystyle n$ be any composite number divisible by $\displaystyle 3$. Let :

$\displaystyle ax^2 + bx + c = n$ for $\displaystyle a, b, c \in \mathbb{Z}$ and $\displaystyle x = 1$.

So, $\displaystyle a + b + c = n$. Furthermore, an additional property is required, which is that $\displaystyle b^2 - 4ac = k^2$, $\displaystyle k \in \mathbb{N}$.

Finally, the last property is that when the polynomial $\displaystyle ax^2 + bx + c = 0$ is factorized into $\displaystyle (mx + p)(m' x + p') = 0$, neither of the factors equal $\displaystyle 3$ when $\displaystyle x = 1$.

I got to the before-last part, because I noticed that :

$\displaystyle b^2 - 4ac = k^2$

$\displaystyle b^2 - k^2 = 4ac$

$\displaystyle (b - k)(b + k) = 4ac$

$\displaystyle b - k = 4a$ and $\displaystyle b + k = c$

So $\displaystyle 2b - 4a = c$

But each time I apply this formula to find $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$, it turns out that the factorized form of the polynomial yields a factor equal to $\displaystyle 3$ when $\displaystyle x = 1$. Why is that ? Why doesn't it yield a nontrivial factor of $\displaystyle n$ ? Does it come from the $\displaystyle 2b - 4a = c$ formula ?

This might not be very clear ... If you need me to make it clearer don't hesitate. Thanks all.