# When $\alpha n + \beta \sqrt{n}$ is an integer for infinitely many n's?

Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R}^2$ with the property that there exist infinitely many positive integers $n$ such that $\alpha n + \beta \sqrt{n}$ belongs to $\mathbb{Z}$.

Note that $S$ trivially contains all the couples $\left( \frac{p}{q}, \frac{r}{\sqrt{q}} \right)$, with $p, q, r$ integers, and $q > 0$. But I don't know whether $S$ contains other elements.

Can we characterize $S$? In particular, is there some $(\alpha, \beta) \in S$, with $\alpha$ irrational?

Yes, $\alpha$ should always be rational. To prove this, suppose that we have $$\alpha n+\beta \sqrt{n}=a$$ and $$\alpha m+\beta\sqrt{m}=b$$ for some integers $a$ and $b$. Solving this system of linear equations in $\alpha$ and $\beta$ we get

$$\beta=\frac{am-bn}{\sqrt{nm}(\sqrt{n}-\sqrt{m})}=\frac{am-bn}{n-m}\left(\frac{1}{\sqrt{m}}+\frac{1}{\sqrt{n}}\right)$$

Therefore, as $\beta\neq 0$ (as otherwise we have trivally $\alpha\in \mathbb Q$), we have for any pair $(n_1,n_2)$ with $\alpha n_i+\beta \sqrt{n_i}\in \mathbb Z$ that $\mathbb Q(\beta)=\mathbb Q(1/\sqrt{n_1}+1/\sqrt{n_2})$. Therefore, $\mathbb Q(\beta)$ is (at most) biquadratic extension. Therefore, if squarefree parts of $n_1$ and $n_2$ are different, we have $\mathbb Q(\beta)=\mathbb Q(\sqrt{n_1}, \sqrt{n_2})$. The latter field has only three quadratic subextensions, so there are only four possibilities for the value of square-free part of $n$ if $\alpha n+\beta\sqrt{n}$ is integer. Thus, there exist $n,m$ with equal squarefree parts and with this property, in which case we have $\mathbb Q(\beta)=\mathbb Q(\sqrt n)$ and

$$\alpha n=\frac{a\sqrt{nm}+bn}{\sqrt{nm}-m}$$

for some integers $a$ and $b$ (this follows by solving the system mentioned before). As the squarefree part of $n$ and $m$ coincide, we have

$$\alpha\in \mathbb Q$$

and

$$\beta\in \sqrt{n}\mathbb Q,$$

which proves both of your assertions.

• Thank you very very much for having answered my question. Your answer is very brilliant, but there are some points unclear to me, many because of some typos. Could you check it, please? Anyhow, I have developed in all details your argument below. It works fine. Thank you very much again! May 10 '18 at 18:07
• Thank you for accepting my answer and for correcting all the typos:) I'm sorry for not being accurate enough. Also, about the possibilities for the value of square-free part: while there are only three quadratic subfields, the square root of $n$ can also generate the subfield $\mathbb Q$, that is, $n$ might also be a perfect square. This, of course, has no effect on the argument, because we only use the fact that $4<\infty$. May 10 '18 at 22:23
• You're right: I had not considered the case in which $n$ is a perfect square! May 11 '18 at 11:14
• I add here a couple of simple comments for those who, like me, are not well acquainted with the theory of field extensions. First of all, if $\gamma=1/ \sqrt{m} + 1/ \sqrt{n}$, let us note that $\mathbb{Q}(\gamma)=\mathbb{Q}(\sqrt{m}, \sqrt{n})$. Clearly $\mathbb{Q}(\gamma) \subseteq \mathbb{Q}(\sqrt{m}, \sqrt{n})$. Now note that $\mathbb{Q}(\gamma)$ must contains $\gamma^2$, and so $\sqrt{mn}$, so also $\delta= \gamma \sqrt{mn}=\sqrt{m}+\sqrt{n}$, and hence $\epsilon= \delta \sqrt{mn}= n \sqrt{m} + m \sqrt{n}$. May 11 '18 at 11:31
• We conclude that $\mathbb{Q}(\gamma)$ also contains $n \delta - \epsilon = (n-m) \sqrt{n}$ and $m \delta - \epsilon = (m -n) \sqrt{m}$ and so $\sqrt{m}$ and $\sqrt{n}$, hence $\mathbb{Q}(\gamma)=\mathbb{Q}(\sqrt{m}, \sqrt{n})$. The second comment is about the fact that if $n$ is a positive integer such that $\alpha + \beta \sqrt{n}$ is an integer, then $n$ can have only four square-free parts. To see this, let $m, n$ be positive integers such that $\alpha n + \beta \sqrt{n}$ and $\alpha m + \beta \sqrt{m}$ are integers. May 11 '18 at 12:40

This is not a separate answer. I only develop here Asymptotiac K's argument, with minor modifications, but including all details, so to make it clear also to people like me who have little knowledge of the theory of field extensions.

We shall prove that $$S = \left\{ \left( \frac{p}{q}, \frac{r}{\sqrt{q}} \right) : p, q, r \in \mathbb{Z}, q > 0 \right\}.$$ First of all, let us remember that for any positive integer $n$, there is a unique representation $n=m^2 k$, where $m$ and $k$ are positive integers, and $k$ is square-free (that is $k$ is divisible by no perfect square other than 1): to see this simply collect all squared factors in the prime factorization of $n$. $k$ is called the square-free part of $n$.

Now suppose that we have for two distinct positive integers $n$ and $m$ such that $$\alpha n+\beta \sqrt{n}=a,$$ and $$\alpha m+\beta\sqrt{m}=b,$$ with $a$ and $b$ integers. Solving this system of linear equations in $\alpha$ and $\beta$ we get

$$\alpha =\frac{a\sqrt{m}-b \sqrt{n}}{n\sqrt{m}-m \sqrt{n}},$$ $$\beta=\frac{bn - am}{n\sqrt{m}-m\sqrt{n}},$$ so that $\alpha$ and $\beta$ belong to $\mathbb{Q}(\sqrt{m},\sqrt{n})$. If $m$ and $n$ have the same square-free part, then from the previous expressions for $\alpha$ and $\beta$ we get that $(\alpha, \beta)= \left( \frac{p}{q}, \frac{r}{\sqrt{q}} \right)$ for some integers $p,q,r$, with $q > 0$.

Now assume that $m$ and $n$ have different square-free parts. Then $\sqrt{m} \notin \mathbb{Q}(\sqrt{n})$ (if for some $\gamma \in \mathbb{Q}$ and $\delta \in \mathbb{Q}$ we had $\sqrt{m}= \gamma + \delta \sqrt{n}$, we should have $m= \gamma^2 + \delta^2 n + 2 \gamma \delta \sqrt{n}$, so that $\gamma=0$ or $\delta=0$, which are both impossible). So $\mathbb{Q}(\sqrt{m}, \sqrt{n})$ is a field extension of degree 4, and $\{1, \sqrt{m}, \sqrt{n}, \sqrt{mn} \}$ is basis of $\mathbb{Q}(\sqrt{m},\sqrt{n})$ considered as a vector space over $\mathbb{Q}$ (see e.g. Artin, Algebra, Chapter 13, Theorem (3.4) and its proof or Birkhoff and Mac Lane, A Survey of Modern Algebra, Chapter 14, Theorem 9). Now consider a positive integer $s$ such that $\alpha s + \beta \sqrt{s}=h$, with $h \in \mathbb{Z}$. Then $\sqrt{s}= (h - \alpha s)/ \beta$, so that $\sqrt{s}$ belongs to $\mathbb{Q}(\sqrt{m},\sqrt{n})$ since $\alpha$ and $\beta$ do. Then there exist unique $\lambda, \mu, \nu, \xi$ such that $$\sqrt{s} = \lambda + \mu \sqrt{m} + \nu \sqrt{n} + \xi \sqrt{mn},$$ and squaring both sides we get $$0 = -s + \lambda^2 + \mu^2 m + \nu^2 n + \xi^2 mn + 2( \lambda \mu + \nu \xi n) \sqrt{m} + 2(\lambda \nu + \mu \xi m ) \sqrt{n} + 2( \lambda \xi + \mu \nu ) \sqrt{mn},$$ and using the fact that $\{1, \sqrt{m}, \sqrt{n}, \sqrt{mn} \}$ is basis of $\mathbb{Q}(\sqrt{m},\sqrt{n})$, we deduce the following system (E) of equations $$\lambda \mu + \nu \xi n = 0,$$ $$\lambda \nu + \mu \xi m = 0,$$ $$\lambda \xi + \mu \nu = 0.$$ Assume that $\lambda \ne 0$. We then get $$\xi = - \frac{\mu \nu}{\lambda},$$ $$\mu ( \lambda^2 - \nu^2 n ) = 0,$$ $$\nu ( \lambda^2 - \mu^2 m ) = 0,$$ and $\mu=0$ or $\nu=0$ would imply $\mu=\nu=\xi=0$, which is impossible. So we should have $\mu^2 m = \lambda^2 = \nu^2 n$, which in turn would imply that $m$ and $n$ have the same square-free part. We conclude that the case $\lambda \ne 0$ is not possible. If instead $\lambda=0$, the system (E) has as solutions $(0,\mu,\nu,\xi)$ where exactly two among $\mu, \nu, \xi$ are equal to zero. Assume e.g. $\nu=\xi=0$. Then, if $\mu= \frac{u}{v}$, with $u$ and $v$ integers, $v^2 s = u^2 m$, so that $s$ and $m$ have the same square-free part. Analogously, if $\mu=\xi=0$, we see that $s$ and $n$ have the same square-free part, while if $\mu=\nu=0$, $s$ and $mn$ have the same square-free part.

Since there are infinitely many positive integers $s$ such that $\alpha s + \beta \sqrt{s}$ is an integer, we can find two of them, let us call them $u$ and $v$ with the same square-free part. Then, if $$\alpha u+\beta \sqrt{u}=c,$$ and $$\alpha v+\beta\sqrt{v}=d,$$ we get $$\alpha =\frac{c\sqrt{v}-d \sqrt{u}}{u\sqrt{v}-v \sqrt{u}},$$ $$\beta=\frac{cv-du}{u\sqrt{v}-v\sqrt{u}},$$ from which we see that for some integers $p, q r$, with $r > 0$ we have $(\alpha, \beta)= \left( \frac{p}{q}, \frac{r}{\sqrt{q}} \right)$.

QED