Find the number of integer solutions of a second degree polynomial equation

[Leo Consoli]

Homework Statement

Being a, b and c integers, find the amount of integer solutions of (x^2 - x - 3 + 2c)/2 = x(ax+b)

Relevant Equations

Girard relations

x^2 - x -3 + 2c = 2x(ax+b)
x^2 -2ax^2 - 2bx - x - 3 + 2c = 0
x^2(1-2a) -x(1+2b) -3 + 2c =0
Using girard
r1+r2 = (1+ 2b)/(1-2a)
r1xr2 = (-3 +2c)/(1-2a)
After this I am stuck.
Thank you.

 

[SammyS]

Problem Statement: Being a, b and c integers, find all integer solutions of (x^2 - x - 3 + 2c)/2 = x(ax+b)
Relevant Equations: Girard relations

x^2 - x -3 + 2c = 2x(ax+b)
x^2 -2ax^2 - 2bx - x - 3 + 2c = 0
x^2(1-2a) -x(1+2b) -3 + 2c =0
Using girard
r1+r2 = 1+ 2b/1-2a
r1xr2 = -3 +2c/1-a
Since the solutions are integers, their sum and their product would also be integers, but after this I am stuck.
Thank you.

You should make the following correct by using parentheses where needed.

Using girard​

r1+r2 = 1+ 2b/1-2a​

r1xr2 = -3 +2c/1-a​

They should read:

r1+r2 = (1+ 2b)/(1−2a)​

r1×r2 = (−3 +2c)/(1−a)​

Those expressions for the sum and product of the roots are not necessarily integers, they're rational expressions in general.

 

[Leo Consoli]

Thank you, I will edit it.

 

[WWGD]

If you add like terms and set to zero, you end up with a second degree polynomial. It's two roots that you can find with the quadratic will be the only possible solutions.

 

[SammyS]

Problem Statement: Being a, b and c integers, find the amount of integer solutions of (x^2 - x - 3 + 2c)/2 = x(ax+b)

So, do you need to find what values a, b, and/or c must take on to get integer solutions, and then determine how many such integer solutions there are?

 

[Leo Consoli]

So, do you need to find what values a, b, and/or c must take on to get integer solutions, and then determine how many such integer solutions there are?

I think that's it.

 

[Leo Consoli]

If you add like terms and set to zero, you end up with a second degree polynomial. It's two roots that you can find with the quadratic will be the only possible solutions.

But can't you make more than one second degree polynomial with the values of a, b and c?

 

[WWGD]

But can't you make more than one second degree polynomial with the values of a, b and c?

You're right, I didn't read carefully, my bad.

 

[SammyS]

Problem Statement: Being a, b and c integers, find the amount of integer solutions of (x^2 - x - 3 + 2c)/2 = x(ax+b)

x^2(1-2a) -x(1+2b) -3 + 2c =0

Rewrite your second degree equation as:
$(2a-1)x^2+(2b+1)x - (2c-3) =0$
Notice that if a, b, and c are all integers, then each of $(2a-1),\,(2b+1),\, (2c-3) \,$ is odd. Also, none of those can be zero, so there are no simple solutions.

Use the master product rule for factoring a second degree polynomial of the form $Ax^2+Bx+C$ where $A,\,B,\,C\,$ are all integers.

 

[Leo Consoli]

In this case, how would I find parcels of AC that add up to B?

 

[SammyS]

In this case, how would I find parcels of AC that add up to B?

(I'm guessing that what you refer to as parcels, I would call factor pairs .)

One clue to answering your question is that A, B, and C are all odd.

 

[Leo Consoli]

That would mean there exist two numbers (m and n) such that
m×n=AC
m+n=B
But AC is odd, and B is also odd, and there isn't a pair of numbers whose sum and product is odd, so its not possible to factorize it, meaning the polynomial has no rational roots.
Is this it?

 

[SammyS]

That would mean there exist two numbers (m and n) such that
m×n=AC
m+n=B
But AC is odd, and B is also odd, and there isn't a pair of numbers whose sum and product is odd, so its not possible to factorize it, meaning the polynomial has no rational roots.
Is this it?

That's the way I see it.

And, if there are no rational roots, then surely there are no integer roots.

 

[Leo Consoli]

Thank you very much for the help.