Are There Two Integer Solutions to This Quadratic Equation?

  • MHB
  • Thread starter anemone
  • Start date
In summary, a quadratic equation is a polynomial equation of the second degree in the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. To determine if it has two integer solutions, the discriminant (b^2 - 4ac) must be a perfect square. It can have a maximum of two integer solutions as the graph of a quadratic equation is a parabola. To solve for integer solutions, the quadratic formula can be used. However, a quadratic equation can have no integer solutions if the discriminant is negative or not a perfect square, resulting in either complex or imaginary solutions.
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
115
Here is this week's POTW:

-----

Let $a,\, b,\, c,\, d \in \mathbb{N} $ such that the equation $x^2-(a^2+b^2+c^2+d^2+1)x+ab+bc+cd+da=0$ has an integer solution. Prove that the other solution is integer too and that both solutions are perfect squares.

-----

 
Physics news on Phys.org
  • #2
Congratulations to Opalg for his correct solution :cool: , which you can find below:

The sum of the two solutions is $a^2+b^2+c^2+d^2+1$, which is an integer. So if one of the solutions is an integer then obviously the other one must also be an integer. Also, the sum and the product of the solutions are positive, so the solutions must both be positive.

Next, $ab \leqslant \tfrac12(a^2+b^2)$ (with equality only if $a=b$), and there are similar inequalities for $bc$, $cd$ and $da$. Therefore $ab+bc+cd+da \leqslant a^2+b^2+c^2+d^2$, with equality only if $a=b=c=d$.

Let $\sigma = ab+bc+cd+da$. Then $a^2+b^2+c^2+d^2 + 1 \geqslant \sigma+1$. So the two solutions of the quadratic equation are positive integers with product $\sigma$ and sum at least $\sigma+1$. That can only happen if the integers are $1$ and $\sigma$ (because if two integers have a given product then their sum is greatest when their difference is greatest). So the sum of the roots is equal to $\sigma + 1$, in other words equality occurs in the above inequality. Therefore $a=b=c=d$, and the quadratic equation becomes $x^2 - (4a^2+1)x + 4a^2$, with solutions $1$ and $4a^2$. Since $1=1^2$ and $4a^2 = (2a)^2$, both solutions are perfect squares.
 

FAQ: Are There Two Integer Solutions to This Quadratic Equation?

1. What is a quadratic equation?

A quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is called quadratic because the highest power of x in the equation is 2.

2. How do you determine if a quadratic equation has two integer solutions?

A quadratic equation will have two integer solutions if its discriminant (b^2 - 4ac) is a perfect square. If the discriminant is positive and a perfect square, the equation will have two distinct integer solutions. If the discriminant is zero, the equation will have one integer solution. If the discriminant is negative, the equation will have no integer solutions.

3. Can a quadratic equation have more than two integer solutions?

No, a quadratic equation can have a maximum of two integer solutions. This is because the graph of a quadratic equation is a parabola, which can intersect the x-axis at most two times.

4. Are there any other methods to solve a quadratic equation besides finding its integer solutions?

Yes, there are several methods to solve a quadratic equation, including factoring, completing the square, and using the quadratic formula. These methods can also be used to find non-integer solutions to a quadratic equation.

5. Can a quadratic equation have no solutions?

Yes, a quadratic equation can have no solutions if its discriminant is negative. This means that the parabola will not intersect the x-axis, and there will be no real solutions to the equation. However, there may still be complex solutions, which involve imaginary numbers.

Back
Top