Math Help - Sum of number as a quadratic equation

1. Sum of number as a quadratic equation

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

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

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

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

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

$b^2 - 4ac = k^2$
$b^2 - k^2 = 4ac$
$(b - k)(b + k) = 4ac$
$b - k = 4a$ and $b + k = c$
So $2b - 4a = c$

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

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

2. Originally Posted by Bacterius
I got to the before-last part, because I noticed that :

$b^2 - 4ac = k^2$
$b^2 - k^2 = 4ac$
$(b - k)(b + k) = 4ac$
$b - k = 4a$ and $b + k = c$
So $2b - 4a = c$
How did you get from line 3 to line 4?

3. Originally Posted by Gusbob
How did you get from line 3 to line 4?
Multiply both expressions from line 4, and you get line 3. There are other ways of doing it such as taking $b - k = 4c$ and $b + k = a$.

4. Try $2x^2+3x+1 = (2x+1)(x+1)$. Is this what you're after?

5. Originally Posted by chiph588@
Try $2x^2+3x+1 = (2x+1)(x+1)$. Is this what you're after?
Well, this still gives a factor of 3 (because 2x + 1 with x = 1), and $2 + 3 + 1 = 6$ which is a composite divisible by 3.

In fact, I'm looking for a systematic way to find $a$, $b$ and $c$ such that $a + b + c = n$ where $n$ is a composite number, and satisfying the condition $b^2 - 4ac$ is a perfect square. Then, when factored, the polynomial $ax^2 + bx + c = 0$ should yield the factorization of $n$ with $x = 1$.

I made it up to the $b^2 - 4ac$ condition, because the only way I can think of that guarantees it to be a perfect square requires n to be a multiple of 3 (and strangely enough, the factorization of the polynomial keeps giving me 3).

So I'm asking if there are any other, and more efficient, methods of finding the values of the coefficients that satisfy both properties.

6. Whoops I thought you wanted an example that had 3, but I guess you want one that doesn't...

Is $5x^2+6x+1=(5x+1)(x+1), \;\; (36-4\cdot5=16=4^2)$ what you're after? I'm a bit confused as to what you want.

Are you saying $b^2-4ac=k^2 \implies 3|n$? If so a counter example is $x^2+4x+3 =(x+3)(x+1)$.

7. No, what I conjecture is that :

$\left. \begin{array}{ll}
(1) \ \ \ b^2 - 4ac &= k^2 \\
(2) \ \ \ 2b - 4a &= c \\
(3) \ \ \ a + b + c &= n
\end{array}
\right \} \implies 3 | n$

(don't know if that's how one formally writes it but I think the idea is pretty straightforward)

And I was wondering if there was a (tractable) way of finding values $a$, $b$ and $c$ that satisfy (1) and (3), but that do not imply that $3 | n$ (or any variant). This question of course assumes that it is (2) that is causing $n$ to be divisible by $3$ (I only conjecture it, again, but I think it can be shown quite easily. If not, we have a new Clay problem anyway xD)

PS : sorry if this is not very clear, I know I really need to work on my communication and logical skills

8. Ok this is much more clear. For your first question let's sub (2) into (3). We get $n=a+b+c=a+b+2b-4a=3b-3a=3(b-a)\implies3|n$.

I'll get back to you on the second part later (I'm in an airport typing on my phone and it's a bit cumbersome ).