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

In summary, there do not appear to be any integer solutions to the equation x^2 - x - 3 + 2c = 2x(ax+b).
  • #1
Leo Consoli
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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.
 
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  • #2
Leo Consoli said:
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.
 
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  • #3
Thank you, I will edit it.
 
  • #4
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.
 
  • #5
Leo Consoli said:
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?
 
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  • #6
SammyS said:
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.
 
  • #7
WWGD said:
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?
 
  • #8
Leo Consoli said:
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.
 
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  • #9
Leo Consoli said:
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.
 
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  • #10
In this case, how would I find parcels of AC that add up to B?
 
  • #11
Leo Consoli said:
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.
 
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  • #12
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?
 
  • #13
Leo Consoli said:
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.
 
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  • #14
Thank you very much for the help.
 

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

1. How do you find the number of integer solutions of a second degree polynomial equation?

To find the number of integer solutions of a second degree polynomial equation, you can use the quadratic formula or factor the equation and set each factor equal to zero. The number of solutions will depend on the discriminant (b^2 - 4ac) of the equation.

2. What is the quadratic formula?

The quadratic formula is a mathematical formula used to solve quadratic equations of the form ax^2 + bx + c = 0. It is written as x = (-b ± √(b^2-4ac)) / 2a.

3. What is the discriminant of a second degree polynomial equation?

The discriminant of a second degree polynomial equation is the expression b^2 - 4ac, which is found in the quadratic formula. It helps determine the number and nature of solutions for the equation.

4. How do you know if a second degree polynomial equation has integer solutions?

A second degree polynomial equation will have integer solutions if the discriminant (b^2 - 4ac) is a perfect square. This means that the square root of the discriminant will be a whole number, resulting in two real and equal solutions.

5. Can a second degree polynomial equation have more than two integer solutions?

Yes, a second degree polynomial equation can have more than two integer solutions. This can happen if the discriminant (b^2 - 4ac) is a perfect square and the equation can be factored into more than two linear factors. Each linear factor will result in an integer solution.

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