 #1
brotherbobby
 421
 107
 Homework Statement:
 Prove that the equation ##x^2+(2m+1)x+(2n+1) = 0## does not possess any rational roots if ##m \in \mathbb{Z}, n \in \mathbb{Z}##.
 Relevant Equations:

1. Discriminant of the quadratic equation ##ax^2+bx+c = 0## is ##\mathscr{D} = b^24ac##
2. If the roots of a quadratic equation are rational, its discriminant must be a perfect square (of an integer), i.e.##\mathscr {D} = p^2## where ##p \in \mathbb{Z}##.
Given : Equation ##x^2+(2m+1)x+(2n+1) = 0## where ##m \in \mathbb{Z}, n \in \mathbb{Z}##, i.e. both ##m,n## are integers.
To prove : If ##\alpha,\beta## be its two roots, then they are not rational numbers.
Attempt : The discriminant of the equation ##\mathscr{D} = (2m+1)^2  4(2n+1) = 4m^2+4m+1  8n  4 = 4m^2+4m(8n+3)##. I need to show therefore that ##4m^2+4m(8n+3) \ne p^2##, where ##p\in \mathbb{Z}##.
If possible, let ##4m^2+4m(8n+3) = p^2##. Being a quadratic equation with an integral root (##m\in \mathbb{Z}##), the solution of this equation ##m = \frac{4 \pm \sqrt{4^2+16(8n+3+p^2)}}{8} = \frac{1\pm \sqrt{4+8n+p^2}}{2}## which implies that the factor within the square root ##4+8n+p^2## must be an odd integer squared, i.e. ##4+8n+p^2 = (2r+1)^2## where ##r \in \mathbb{Z}##.
Rearranging, we get ##p^2 = 4r^2+4r8n3##. Is the expression on the right a perfect square, for the sake of p?
I find myself stuck here. A help would be welcome.
To prove : If ##\alpha,\beta## be its two roots, then they are not rational numbers.
Attempt : The discriminant of the equation ##\mathscr{D} = (2m+1)^2  4(2n+1) = 4m^2+4m+1  8n  4 = 4m^2+4m(8n+3)##. I need to show therefore that ##4m^2+4m(8n+3) \ne p^2##, where ##p\in \mathbb{Z}##.
If possible, let ##4m^2+4m(8n+3) = p^2##. Being a quadratic equation with an integral root (##m\in \mathbb{Z}##), the solution of this equation ##m = \frac{4 \pm \sqrt{4^2+16(8n+3+p^2)}}{8} = \frac{1\pm \sqrt{4+8n+p^2}}{2}## which implies that the factor within the square root ##4+8n+p^2## must be an odd integer squared, i.e. ##4+8n+p^2 = (2r+1)^2## where ##r \in \mathbb{Z}##.
Rearranging, we get ##p^2 = 4r^2+4r8n3##. Is the expression on the right a perfect square, for the sake of p?
I find myself stuck here. A help would be welcome.