Find two numbers that satisfy these conditions

**Bacterius** · February 12th 2010, 03:15 AM

Hi all,
this is a question I've been doing. Let $n$ (known) be some composite odd integer. Find two numbers $b$ and $c$ such that :
- $x^2 + bx + c + n = 0$ is factorable over the real numbers, with integer coefficients in the standard form and factorized form.
- $x^2 + bx + c = 0$ 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 $\Delta_1$ and $\Delta_2$ denote the discriminants of equations one and two. We have :

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

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

And so, if I could find the values of $\Delta_1$ and $\Delta_2$ , I could easily work out $b$ and $c$ 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

**Soroban** · February 12th 2010, 07:51 AM

Hello, Bacterius!

I've one observation so far . . .

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

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

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

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

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

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

**Bacterius** · February 12th 2010, 01:22 PM

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 $(c + n)$ but $(c - n)$ in the quadratics and the discriminants, however it does not change your solution. But what if the factorization of $n$ is unknown ?

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

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

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

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

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

**Bacterius** · February 13th 2010, 09:17 PM

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

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

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

**SuperCalculus** · February 14th 2010, 01:54 AM

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.

**Bacterius** · February 14th 2010, 02:04 AM

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 $0 = 0$ .

**SuperCalculus** · February 14th 2010, 02:10 AM

Originally Posted by Bacterius

I know how you feel, having fiddled with algebra for ten minutes just to find that $0 = 0$ .

Or to find that $x^2 = y^2 + 4n$ for the fifteenth time.

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