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.