- #1

brotherbobby

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- 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^2-4ac##

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+4r-8n-3##. Is the expression on the right a perfect square, for the sake of

*p*?

**I find myself stuck here. A help would be welcome.**