# Question regarding the roots of $x^2 + nx + m$, for $n,m\in\mathbb{Z}$.

The equation $$x^2+nx+m=0$$, $$n,m\in\mathbb Z$$, cannot have

A. Integral roots
B. Non-integral roots
C. Irrational roots
D. Complex roots

One thing you can try to do for a problem like this is write down some explicit values of $$(m,n)$$, and see what happens!

How I'd approach this problem in particular is try to craft equations that have the roots that we want to show it can't have.

A. Integral roots:

We want $$x^2 + nx + m$$ to have integral roots. One easy root to give it is $$x = 0$$, by setting $$m = 0$$. $$x^2 + nx = x(x+n)$$ has roots $$x,-n$$, for any choice of $$n$$. So A clearly isn't true!

B. Non-integral roots

You can try this for various values, and you might struggle to find rational roots. This doesn't prove that this is the answer, but it's good to note for sure.

C. Irrational roots

If we set $$n = 0$$, we get the equation $$x^2 + m = 0$$, and by choosing $$m < 0$$ we can $$\sqrt{k}$$ as a root (for any $$k$$ we want!). For example, $$x^2 -2$$ has root $$\sqrt{2}$$. This clearly isn't the choice then.

D. Imaginary roots

If we again set $$n = 0$$, but instead choose $$m > 0$$, we'll get imaginary roots! We can use the equation $$x^2 + 1$$, to get $$i$$ as a root!

The above makes it seem like B. should be the answer. We might then wonder why $$B$$ should be the answer. Someone else already mentioned the Rational Root test, which gives an easy answer as to why. A more qualitative way to see this is by trying to build an equation of your form that has a rational root!

Let the two roots of our equation be $$r_1 = p/q$$, and $$r_2$$ which can be anything. Then, our equation is:

$$(x - p/q)(x - r_2) = 0 \implies (qx-p)(x-r_2) = 0 \implies qx^2 - (r_2q +p)x + pr_2 = 0\implies x^2 - \frac{r_2q+p}{q}x + \frac{pr_2}{q} = 0$$

We now just need $$(r_2q+p)/q$$ and $$pr_2/q$$ to be integral. We can "force" the second one to be integral by making $$r_2$$ a multiple of $$q$$ (say $$r_2 = kq$$ for some integer $$k$$). But then we have that $$(kq^2+p)/q$$ needs to be an integer, which it won't be.

If we instead want to force $$(r_2q+p)/q$$ to be an integer, we could say that we'll set $$r_2$$ such that $$r_2q+p = kq$$ for some integer $$k$$. This is the same as saying that $$p = (k-r_2)q$$, which unfortunately isn't allowed at first (While I didn't explicitly say so, generally when you write a rational number as $$p / q$$ you assume that $$p$$ and $$q$$ are coprime, so have no common factors. The equation I just wrote down shows that $$q$$ is a common factor, which would imply that $$p / q$$ was integral all along).

• Non-integral roots include irrational roots, which is why I consider this a poor question. Please see my answer. – Deepak Jan 15 at 6:43

The entire nature of the roots depends upon auxiliary equation ie.,$$n^2-4m$$ let us call it as AE Case-1$$AE >0$$ the root maybe an integer or a surd ex:-$$(x^2-3x+2)(1,2),(x^2-3x+1)(1{^+_-}\frac{\sqrt5}2)$$ case-2if$$AE =0$$ the roots are integer and equal ex:-$$(x^2+2x+1)(1,1)$$ case-3 if $$AE <0$$ roots are imaginary and complex conjugates to each other ex:- $$(x^2-2x+3)(1{^+_-}i\sqrt2),(x^2-2x+5)(1{^+_-}i2),(x^2+1)({^+_-}i).$$

A. $$n=2, m=1$$ is a counterexample.
C. $$n=1, m=-1$$ is a counterexample.
D. $$n=m=1$$ is a counterexample.
To see why this is the case, consider the solution by the general quadratic formula $$x = \frac 12(-n\pm \sqrt{n^2-4m})$$. We are considering only cases where $$n^2-4m$$ is a perfect square, otherwise the roots would be irrational. Now $$n$$ has to be either even or odd. If $$n$$ were even, $$\sqrt{n^2-4m}$$ would be even as would $$-n\pm \sqrt{n^2-4m}$$ so the roots would be integral. If $$n$$ were odd, then $$\sqrt{n^2-4m}$$ would be odd and hence $$-n\pm \sqrt{n^2-4m}$$ would be even, again rendering the roots integral. There is no way to construct non-integral rational roots.