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Prove that if $a,b \in \mathbb{Q}$, and $a+b \in \mathbb{Z}$, $2ab \in \mathbb{Z}\,$ [and $\color{#0a0}{a^2+b^2} \in \mathbb{Z}$] , then $a, b \in \mathbb{Z}.$

[N.B. the original question did not include the hypothesis $\,\color{#0a0}{a^2+b^2}\in \Bbb Z.\,$ This is redundant, since it is equivalent to $\,2ab\in\Bbb Z,\,$ assuming $\,a+b\in\Bbb Z,\,$ by using $\,(a+b)^2 = \color{#0a0}{a^2+b^2}+ 2ab\ $ -Bill D.]

I used the Rational root theorem on polynomial $x^2 - (a+b)x + 2ab$

All possible roots of the polynomial are divisors of 2ab and are integers. But then using quadratic formula $x_{1,2}=a+b\pm\frac{\sqrt{(a+b)^2-8ab}}{2}$ Then, because roots are integers $(a+b)^2-8ab$ has to be an even perfect square, I do not know what to do now.

Bill Dubuque
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mathchad
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    What have you tried till now? – Dstarred Jan 10 '23 at 14:59
  • Using rational root theorem in polynomial x^2 -(a+b) + 2ab , and similar ones, but i didnt get anything interesting – mathchad Jan 10 '23 at 15:00
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    Please edit your question to show your attempt. This MathJax tutorial explains how to typeset mathematics on this site. – N. F. Taussig Jan 10 '23 at 15:02
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    The last condition is redundant – nonuser Jan 10 '23 at 15:03
  • @nonuser I think you are saying $a^2+b^2 \in \mathbb Z$ is redundant given the other two conditions, but since the title and the text have changed, not always consistently, it might be worth being clear – Henry Jan 10 '23 at 15:10
  • Hint: 1) show that $2a, 2b$ are integers using RRT (modifying your polynomial slightly - Note that the roots of the current polynomial are not easily known as yet) 2) Hence show that $a, b$ are integers (since $2ab$ is an integer). – Calvin Lin Jan 10 '23 at 15:50
  • By the linked dupe, in least terms $a$ and $b$ have the same denominator, which must be $1$ if $2ab\in \Bbb Z.,$ You can also do it using 2-adic valuations $,\nu_2,$ and various related ways, but all will be dupes. – Bill Dubuque Jan 10 '23 at 16:49
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    What kind of dupe is this? If this question is deleted, it cannot be found in the search results and therefore the relevant link cannot be followed. I'm voting to open this question. – lone student Jan 10 '23 at 17:00
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    Dupe of this. Was wrongly reopened after the title was edited to obscure the dupe relationship. It is an immediate consequence of the basic fact proved there - if a sum of rationals is an integer then they have equal least denominators. The dupe proves a result a result more general than this question (replacing $2ab$ by $nab$ for squarefree $n$, with the same simple proof sketched above). – Bill Dubuque Jan 10 '23 at 19:57
  • @lonestudent That's not what i was thinking. For part 2, we do still need the fact that $a+b \in \mathbb{Z}$, which I didn't explicitly state. It is similar to Bill's dupe (and works for square-free $n$). – Calvin Lin Jan 10 '23 at 21:46
  • @CalvinLin Yes. I got it. You replied after deleting the comment :-) – lone student Jan 10 '23 at 21:56
  • @CalvinLin I added a lower bound tweak to RRT to completely handle the $n$-ary version of this - see this Corollary (probably something like that is already in the contest community toolbox). – Bill Dubuque Jan 12 '23 at 00:43
  • Dupe 2 This is a special case of a (tweaked) Rational Root Test - see the prior comment. – Bill Dubuque Jan 12 '23 at 13:36

1 Answers1

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Let $\thinspace a+b=m,\thinspace 2ab=n$, where $m,n\in\mathbb Z\thinspace .$ Then we have:

$$ \begin{align}&\begin{cases} 2a(m-a)-n=0\\ 2b(m-b)-n=0\end{cases}\\\\ \implies &\begin{cases}2a^2-2am+n=0\\ 2b^2-2bm+n=0\end{cases}\end{align} $$

and

$$\begin{align}\Delta_a=\Delta_b &=m^2-2n\\ &=k^2,\thinspace k\in\mathbb Z\thinspace. \end{align}$$

This yields,

$$ \begin{align}a=\frac {m\pm k}{2},\thinspace \thinspace \thinspace b=\frac {m\mp k}{2}\end{align} $$

If $a,b\in\mathbb Q\setminus \mathbb Z$, then $m$ and $k$ have opposite parities.

This leads to:

$$n=\frac{m^2-k^2}{2}\not\in\mathbb Z$$

A contradiction.

lone student
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  • Simpler: $A,B = 2a,2b$ are roots of $f(x) = (x!-!A)(x!-!B)=x^2-2mx+2n,$ so $A,B\in \Bbb Z,$ by RRT, so $f(A)!=!0!=!f(B)\Rightarrow A^2,B^2$ even $\Rightarrow A,B$ even $\Rightarrow ,a,b\in \Bbb Z.\ \small\bf QED$. Using the quadratic formula as you do only obscures these simple basic parity inferences. [This way too is a dupe] $\ \ $ – Bill Dubuque Jan 10 '23 at 21:41
  • Your answer seems fine. Although, as Bill pointed out as well, you didn't really need the QF here to establish opposite parities – 冥王 Hades Jan 10 '23 at 22:08
  • @Gokuカカロット The method used in the answer is known as Proof by Contradiction. Therefore, in general, technique has to be used in a way that leads to a contradiction. – lone student Jan 10 '23 at 22:44
  • I added a lower bound tweak to RRT to allow it to handle the degree $n$ generalization of this - see the Corollary here. Of course generally we can't solve arbitrary degree polynomials as in the above answer, but we don't need to - we only need (tweaked) RRT, and it yields a one-line proof. – Bill Dubuque Jan 12 '23 at 00:46