# Thread: Find two numbers that satisfy these conditions

1. ## Find two numbers that satisfy these conditions

Hi all,
this is a question I've been doing. Let $\displaystyle n$ (known) be some composite odd integer. Find two numbers $\displaystyle b$ and $\displaystyle c$ such that :
- $\displaystyle x^2 + bx + c + n = 0$ is factorable over the real numbers, with integer coefficients in the standard form and factorized form.
- $\displaystyle x^2 + bx + c = 0$ is factorable over the real numbers, with integer coefficients in the standard form and factorized form.

I've been thinking, for a quadratic equation to satisfy these properties, its discriminant must be positive and be a perfect square. Let $\displaystyle \Delta_1$ and $\displaystyle \Delta_2$ denote the discriminants of equations one and two. We have :

$\displaystyle \Delta_1 = b^2 - 4(c + n)$
$\displaystyle \Delta_2 = b^2 - 4c$

And I also noted that $\displaystyle \Delta_1 - \Delta_2 = b^2 - 4(c + n) - b^2 + 4c = 4c - 4(c + n) = 4(c - c + n) = 4n$, and this could be useful regarding the difference of two squares.

And so, if I could find the values of $\displaystyle \Delta_1$ and $\displaystyle \Delta_2$, I could easily work out $\displaystyle b$ and $\displaystyle c$ with a simple system of equations. But what I've done beyond this point does not lead to anything interesting, so I'm wondering, do I have enough information about the discriminants to work them out ?

Does anyone have any idea to put me on the right track again ? Thanks all

2. Hello, Bacterius!

I've one observation so far . . .

Let $\displaystyle n$ (known) be some composite odd integer.
Find two numbers $\displaystyle b$ and $\displaystyle c$ such that :

$\displaystyle x^2 + bx + c + n \:=\: 0$ is factorable over the real numbers, with integer coefficients.

$\displaystyle x^2 + bx + c \:=\: 0$ is factorable over the real numbers, with integer coefficients.

$\displaystyle \text{Since }n\text{ is composite, let one factoring be: }\:n \:=\:p\cdot q$

$\displaystyle \text{Let }c = 0\,\text{ and }\,b \:=\:p+q$

$\displaystyle \text{Then both }\:\begin{Bmatrix}x^2 + (p+q)x + pq \\ \\[-3mm] x^2 + (p+q)x\end{Bmatrix}\:\text{ are factorable over the real numbers.}$

3. Thank you Soroban, I didn't see this particular case, thanks. Actually, I made a little sign mistake in my original post, it isn't $\displaystyle (c + n)$ but $\displaystyle (c - n)$ in the quadratics and the discriminants, however it does not change your solution. But what if the factorization of $\displaystyle n$ is unknown ?

For example, let $\displaystyle n = 33$. I want to find two numbers $\displaystyle b$ and $\displaystyle c$ that satisfy :
- $\displaystyle b^2 - 4c$ is a perfect square
- $\displaystyle b^2 - 4(c - n)$ is a perfect square

After some guess and check, I find that two numbers that work are $\displaystyle b = 2$ and $\displaystyle c = -15$, because :

$\displaystyle 2^2 - 4 \times (-15) = 64 = 8^2$
$\displaystyle 2^2 - 4 \times (-15 - 33) = 196 = 14^2$

And thus, $\displaystyle x^2 + 2x - 15$ is factorable over the reals with integer coefficients, as well as $\displaystyle x^2 + 2x - 48$.

But there is a problem : I used guess and check. That's why I was looking for a particular process to find $\displaystyle b$ and $\displaystyle c$, more efficient than trial.

4. Ok, so this is the actual problem, reformulated in a easier to understand yet equivalent way :

I am looking for two integers, $\displaystyle x$ and $\displaystyle y$.
I know the value of some integer $\displaystyle n$, that can vary.
I also know that $\displaystyle x^2 = b^2 - 4(c - n)$ and $\displaystyle y^2 = b^2 - 4c$ for some integers $\displaystyle b$, $\displaystyle c$.
I also know that $\displaystyle x^2 - y^2 = b^2 - 4(c - n) - (b^2 - 4c) = b^2 - 4c + 4n - b^2 + 4c = 4n$.

Do I have enough information to find $\displaystyle x$ and $\displaystyle y$ ? Or do I need some other information like their sum or product ?

5. Argh asdf this is annoying. It seems to be that you need to use some form of substitution, but everything I try seems to go around in circles.

6. Originally Posted by SuperCalculus
Argh asdf this is annoying. It seems to be that you need to use some form of substitution, but everything I try seems to go around in circles.
I know how you feel, having fiddled with algebra for ten minutes just to find that $\displaystyle 0 = 0$.

7. Originally Posted by Bacterius
I know how you feel, having fiddled with algebra for ten minutes just to find that $\displaystyle 0 = 0$.
Or to find that $\displaystyle x^2 = y^2 + 4n$for the fifteenth time.