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the question

Show that if $a$ and $q$ are natural numbers and the number $(a+\sqrt{q})(a+\sqrt{q+1})$ is rational, then $q=0$.

the idea

for the number to be rational both members have to be rational (*)

because a is natural, it means that $\sqrt{q}$ and $\sqrt{q+1}$ should both be perfect squares, but they are also consecutive

$\sqrt{q}+1=k^2+1$ => $\sqrt{q+1}=k^2+2k+1$

The equality would happen only id $2k=0 => k=0 => q=0$

Im not sure of the part I noted (*), because I think I should also demonstrate this fact, but I don't know how. Hope one of you can tell me if my idea is right and how can I improve my answer! Thank you!

IONELA BUCIU
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    $(*)$ isn't true (in generality, that is. It is true here, but that's only after one sees that $q=0$ is the only possibility). – Sarvesh Ravichandran Iyer Feb 12 '24 at 16:46
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    I'm supposing your definition of natural numbers includes $0$? – D S Feb 12 '24 at 16:54
  • It's possible for the product of two distinct irrational numbers to be rational. For example, $(2 + \sqrt 2)(2 - \sqrt 2) = 2 \in \mathbb{Q}$. – Dan Feb 12 '24 at 16:58
  • I agree with Dan's comment. It's not simply a matter of assuming $\sqrt q$ and $\sqrt{q+1}$ are both rational (if that were all whats the $a$ for and why word it this weird way) but either $\sqrt q, \sqrt{q+1}$ are both rational (and $q=0$) or both irrational. If $a+\sqrt{q}$ is irrational but $(a+\sqrt{q})(a+\sqrt{q+1})$ is rational then $a+\sqrt{q+1} = r \frac 1{a+\sqrt q}$ for some rational $r$. You can prove that is impossible. (probably more easily than I just set it up for) – fleablood Feb 12 '24 at 17:55

4 Answers4

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Let $x:=(a+\sqrt q)(a+\sqrt{q+1})$.

  • If $x=0$ then $a=q=0$.

  • If $x\ne0$, we can borrow the first step of @FredBernard's answer: if $x$ is rational, so is $y:=(a-\sqrt{q})(a-\sqrt{q+1})$, because $xy=(a^2-q)(a^2-q-1)$ is an integer.
    Then, $\sqrt{q(q+1)}=\frac{x+y}2-a^2\in\Bbb Q$ hence $q(q+1)$ is a perfect square, hence so are its two coprime factors $q$ and $q+1$, but $m=\sqrt q,n=\sqrt{q+1}\implies1=(n-m)(n+m)\implies n=1,m=0\implies q=0$.

Anne Bauval
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$x=(a+\sqrt{q})(a+\sqrt{q+1})$ is rational iff $y=(a-\sqrt{q})(a-\sqrt{q+1})$ is, because $xy$ belongs to $\mathbb Z$. Hence in this case $x+y$ is rational i.e $\sqrt{q}\sqrt{q+1}$ is. Then for some integers $\alpha,\beta$ with $GCD(\alpha,\beta)=1$ we have $\beta^2q(q+1)=\alpha^2$ so if $\alpha\neq 0$, $\alpha^2$ is a divisor of $q(q+1)$ and we have $1=\beta^2\lambda$ where $\lambda=\frac{q(q+1)}{\alpha^2}$. This entails $\beta=\lambda=1$ and $q(q+1)=\alpha^2$ so $\alpha^2-q^2=q$ : this is not possible as $\alpha$ would be then greater than $q+1$ and thus $\alpha^2-q^2\geqslant 2q+1>q$. So $\alpha$ and hence $q$ must be zero.

Fred Bernard
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  • This is nice (+1). You can improve the ending by $\sqrt{q^2+q} \in \Bbb Q \implies \sqrt{q^2+q} \in \Bbb Z$, as this is well known. But $q^2 \le q^2+q< q^2+2q+1$, hence the result follows. – D S Feb 12 '24 at 18:20
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If $a \neq 0$:
FTSOC, assume both $\sqrt{q}$ and $\sqrt{a+1}$ are irrational (if one is rational, then so is the other). Then $a \neq \sqrt q$ and: $$(a+\sqrt{q})(a+\sqrt{q+1}) \in \Bbb Q$$ $$\iff a(\sqrt q+\sqrt{q+1})+\sqrt{q^2+q} \in \Bbb Q \tag{1}\label{1}$$ Also: $$\frac{a-\sqrt{q}}{a-\sqrt q}\cdot (a+\sqrt{q})(a+\sqrt{q+1}) \in \Bbb Q$$ $$\iff \frac{a+\sqrt{q+1}}{a-\sqrt q} \in \Bbb Q$$ $$\iff \frac{(a-\sqrt q)+\sqrt q+\sqrt{q+1}}{a-\sqrt q} \in \Bbb Q$$ $$\iff \frac{\sqrt q+\sqrt{q+1}}{a-\sqrt q} \in \Bbb Q$$ $$\iff \frac{1}{(a-\sqrt q)(\sqrt{q+1}-\sqrt q)} \in \Bbb Q$$ $$\iff a\sqrt{q+1}+q - a\sqrt{q}-\sqrt{q^2+q} \in \Bbb Q$$ $$\iff 2a\sqrt{q+1} - [a(\sqrt q+\sqrt{q+1})+\sqrt{q^2+q}] \in \Bbb Q$$ From $\eqref{1}$, we have $$\iff 2a\sqrt{q+1} \in \Bbb Q$$ $$\iff \color{red}{\sqrt{q+1} \in \Bbb Q}$$

If $a=0$, then $\sqrt{q^2+q}$ is rational, and this implies it is an integer. But $q^2 \leq q^2+q<q^2+2q+1 \implies q^2+q = q^2 \implies q = 0$.

D S
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This is a "reader's digest" of the magic main part of @DS's answer.

Assume $a>0$, and let $$\alpha:=a+\sqrt q,\quad\bar\alpha:=a-\sqrt q,\quad\beta:=a+\sqrt{q+1}.$$ Note that:

  • $\alpha(\beta-\bar\alpha)\ne0$
  • $\alpha\bar\alpha=a^2-q\in\Bbb Q$
  • $(\beta-\bar\alpha)(\beta-\alpha)=1\in\Bbb Q$
  • $(\alpha\beta-\alpha\bar\alpha)(\bar\alpha\beta-\alpha\bar\alpha)=\alpha\bar\alpha(\beta-\bar\alpha)(\beta-\alpha)$.

Therefore, if $\alpha\beta\in\Bbb Q$ then $\bar\alpha\beta\in\Bbb Q$, hence $\beta=\frac{\alpha\beta+\bar\alpha\beta}{2a}\in\Bbb Q$.

Anne Bauval
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