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This is an exercise handled us by the teacher on a first year math-degree.

The question reads:

Prove that if n,m are odd integers, then the equation $x^2 +2mx+2n=0$ has no rational solutions. Hint: Trying by divisibility, proof first that if it has rational solutions they have to be integers, and then prove that it can't have integer solutions.

My solution is the following:

A Suppose that $x^2+2mx+2n=0$ has a rational solution. Then we can write x=p/q, p/q are coprimes integers, so that x is irreductible. Then, if we substitute: $$(p/q)^2+2m(p/q)+2n=0 \Rightarrow (p/q)^2+2m(p/q)=-2n $$ That tells us that, because $-2n$ has to be an integer, then so has to be the left side. If we rearrenge the left side we have $$\frac{p^2+2mpq}{q^2}=\frac{p(p+2mq)}{q^2}$$ We know that it has to be an integer, so either $q^2 \mid p$ or $q^2 \mid (p+2mq)$. The first one can't be, so $q^2 \mid (p+2mq)$ must be true. If that is true, $$\frac{q(p/q+2m)}{q^2} \quad \textrm{simplifying} \quad \frac{p/q+2m}{q}=k$$ $$\frac{p}{q}+2m=kq$$ That can't be because 2m is an integer, and p/q a rational, and a integer+rational can't equal an integer. That means that $p/q$ can't exist if they don't equal an integer, so if $x$ is a rational, it must be an integer.

B Now we are going to prove that x can't be an integer. For that, we are going to use the fact that $m,n$ are odd.

  1. Case $1$: $x$ is odd

If $x$ is odd, then $(2k+1)^2+2(m(2k+1))+2n=0 \Rightarrow (2k+1)^2+2(m(2k+1))= -2n$. This can't be because $(2k+1)^2$ is odd, and odd+even=odd, but $-2n$ must be and even integer.

  1. Case $2$: $x$ is even.

If $x$ is even, then $4k^2+4mk+2n=0 \Rightarrow 2k^2+2mk=-n$, which is also impossible, because $-n$ is odd, and the left side must be an even integer because even+even=even.

With that, we conclude that x can't be an integer, so there are no solutions x such that x is rational. The problem with my solution is that it doesn't use the fact that $m$ is odd, which usually means that I'm overlooking something.

Is the solution correct? Thanks.

insipidintegrator
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  • We can drastically simplify this : $(1)$ If there is a rational solution , it must be an integer dividing $2n$ , so there there are two integer solutions multiplying to $2n$. $(2)$ Since $n$ is odd , exactly one of those integers must be even, the sum is then odd contradicting Vieta's formula. – Peter Oct 09 '22 at 14:21
  • Easier: it is irreducible over $\Bbb Q$ by Eisenstein's criterion for $,p=2\ \ $ – Bill Dubuque Oct 09 '22 at 14:22
  • That you didn't use all the hypotheses implies nothing at all about the correctness of your solution. Here that hypothesis is not needed. For SV questions to be on topic you need to specify precisely what step of the proof you doubt, and why so. See the linked dupe (and its links) for many methods of solution. – Bill Dubuque Oct 09 '22 at 14:42
  • In A there is an unjustified claim that $q^2 \mid p$ or $q^2 \mid (p+2mq)$. More simply, clearing denoms yields $0 = q^2f(p/q) = p^2+q(\ldots)$ so $,q\mid p^2,,$ contra $(q,p)=1$. This proof is a special quadratic case of a proof of the Rational Root Test. The argument in B can be simplified in many ways, eg.. $,\color{#c00}{x(x+2m)}=2n\Rightarrow 2\mid x^2\Rightarrow 2\mid x \Rightarrow 4\mid {\rm\color{#c00}{ LHS}}\Rightarrow 4\mid {\rm RHS}!=!2n\Rightarrow 2\mid n,,$ contra hypothesis. – Bill Dubuque Oct 09 '22 at 15:22

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