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#1 h Apr 18, 2025, 7:46 PM

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Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers $a$ and $b$ , there cannot exist infinitely many $n \in \mathbb{Z}$ such that
$\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.$
$\textbf{Solution}$

Let
$x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.$
Then,
$x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.$ So:
$x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.$
Therefore,
$\sqrt{(n + a)(n + b)} \in \mathbb{N}.$
Let
$(n + a)(n + b) = k^2.$ Assume $n + a \neq n + b$ . Then we have:
$n + a \mid k \quad \text{and} \quad k \mid n + b,$ or it could also be that $k \mid n + a \quad \text{and} \quad n + b \mid k$ .

Without loss of generality, we take the first case:
$(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.$
Thus,
$k_1 k_2 = \frac{n + b}{n + a}.$
Since $k_1 k_2 \in \mathbb{N}$ , we have:
$k_1 k_2 = 1 + \frac{b - a}{n + a}.$
For infinitely many $n$ , $\frac{b - a}{n + a}$ must be an integer, which is not possible.

Therefore, there cannot be infinitely many such $n$ .

This post has been edited 2 times. Last edited by Levieee, Apr 18, 2025, 7:47 PM

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#2 h Apr 18, 2025, 8:10 PM

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I might be missing something, but why does $(n+a)(n+b)=k^2$ imply $n+a\mid k$ and $k\mid n+b$ ? Why cant for example $n+a=25$ and $n+b=4$ and $k=10$ . The factors of $k$ do not have to be distributed nicely between $n+a$ and $n+b$ .

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#3 h Apr 18, 2025, 8:45 PM

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I think the condition implies both $n+a$ and $n+b$ are perfect squares.

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#4 h Apr 18, 2025, 9:04 PM • 1 Y

Y by Levieee

You need a slightly more sophisitcated finish after realising $(n+a)(n+b)$ is a square for infinitely many $n$ and it is that we have $n^2+(a+b)n+ab$ is a square for infinitely many $n$ and now setting some arbitrarily large $n$ and considering some $k$ for which $2k \ge a+b >2k-2$ we could have that it is a square bounded between $(n+k)^2$ and $(n+k-1)^2$ so it has to be $(n+k)^2$ which shows that $(a+b)^2=4ab$ by coefficient checking and matching and thus $(a-b)^2=0$ so $a=b$ which is a contradiction!.

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Safal

#5 h Apr 18, 2025, 9:18 PM

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Levieee wrote:

Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers $a$ and $b$ , there cannot exist infinitely many $n \in \mathbb{Z}$ such that
$\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.$
$\textbf{Solution}$

Let
$x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.$
Then,
$x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.$ So:
$x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.$
Therefore,
$\sqrt{(n + a)(n + b)} \in \mathbb{N}.$
Let
$(n + a)(n + b) = k^2.$ Assume $n + a \neq n + b$ . Then we have:
$n + a \mid k \quad \text{and} \quad k \mid n + b,$ or it could also be that $k \mid n + a \quad \text{and} \quad n + b \mid k$ .

Without loss of generality, we take the first case:
$(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.$
Thus,
$k_1 k_2 = \frac{n + b}{n + a}.$
Since $k_1 k_2 \in \mathbb{N}$ , we have:
$k_1 k_2 = 1 + \frac{b - a}{n + a}.$
For infinitely many $n$ , $\frac{b - a}{n + a}$ must be an integer, which is not possible.

Therefore, there cannot be infinitely many such $n$ .

$k_1$ and $k_2$ need not be integers, they can be rational beacuse we can have that the denominator ot that rational divide $n+a$

This post has been edited 1 time. Last edited by Safal, Apr 18, 2025, 9:19 PM

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233 posts

#6 h Apr 19, 2025, 7:39 AM

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Safal wrote:

Levieee wrote:

Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers $a$ and $b$ , there cannot exist infinitely many $n \in \mathbb{Z}$ such that
$\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.$
$\textbf{Solution}$

Let
$x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.$
Then,
$x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.$ So:
$x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.$
Therefore,
$\sqrt{(n + a)(n + b)} \in \mathbb{N}.$
Let
$(n + a)(n + b) = k^2.$ Assume $n + a \neq n + b$ . Then we have:
$n + a \mid k \quad \text{and} \quad k \mid n + b,$ or it could also be that $k \mid n + a \quad \text{and} \quad n + b \mid k$ .

Without loss of generality, we take the first case:
$(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.$
Thus,
$k_1 k_2 = \frac{n + b}{n + a}.$
Since $k_1 k_2 \in \mathbb{N}$ , we have:
$k_1 k_2 = 1 + \frac{b - a}{n + a}.$
For infinitely many $n$ , $\frac{b - a}{n + a}$ must be an integer, which is not possible.

Therefore, there cannot be infinitely many such $n$ .

$k_1$ and $k_2$ need not be integers, they can be rational beacuse we can have that the denominator ot that rational divide $n+a$

yea but after it's reduced we can get it to be an integer , doesnt $a \mid b$ imply for some $k$ $ak=b$ , k is an integer

@2above how do u say that $n^2 + ab + (a+b)n$

This post has been edited 2 times. Last edited by Levieee, Apr 19, 2025, 7:41 AM

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#7 h Apr 19, 2025, 7:42 AM

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DTforever wrote:

I think the condition implies both $n+a$ and $n+b$ are perfect squares.

does the counter example provided by @Lil_flip38 work? or that counter example can be invalidated because $n = 4-b$ , $b \ge 1$ so as n increases we can just show it cant happen

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#8 h Apr 19, 2025, 8:33 AM • 1 Y

Y by Levieee

Levieee wrote:

Assume $n + a \neq n + b$ . Then we have:
$n + a \mid k \quad \text{and} \quad k \mid n + b,$ or it could also be that $k \mid n + a \quad \text{and} \quad n + b \mid k$ .

This part is certainly wrong, since $(n+a)(n+b)=k^2$ can imply $n+a \leq k$ and $n+b \geq k$ or vice-versa but not the divisibility condition above. You can find many counter examples to this.

Another finish from this part can be that let $\gcd(n+a,n+b)=d$ . (of course, $n+a \neq n+b$ )
So, $n+a = dx$ and $n+b=dy$ , $\gcd(x,y)=1$
This gives us that, $d^2xy = k^2$
Which means that $xy=l^2$ and $x=u^2$ and $y=v^2$ , $\gcd(u,v)=1$ .
This means that, $b-a=d(v^2 - u^2)$ for a particular $n$ .
Since, the original condition needs to be true for infinitely many $n$ , by infinite PHP we will have infinitely many $n$ for which we get the same value of $d$ for a particular value of $k$ .
So, for infinitely many $n$ , which in turn give infinitely many value of the pair $(u,v)$ , the value $\frac{b-a}{d}$ is constant, say $t$ .
From the above argument, since infinitely many pairs $(u,v)$ satisfy $t = v^2-u^2 = (v-u)(v+u)$ , where $v,u$ are integers, this means that $t$ has infinitely many factors $\implies$ $t=0$ , i.e., $b=a$ which contradicts that $a$ and $b$ are distinct.

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