Y by
Idk where it went wrong, marks was deducted for this solution

Show that for a fixed pair of distinct positive integers
and
, there cannot exist infinitely many
such that
![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](//latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)

Let
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](//latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
Then,
So:
![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](//latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
Therefore,
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](//latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
Let
Assume
. Then we have:
or it could also be that
.
Without loss of generality, we take the first case:
![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](//latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
Thus,
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](//latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
Since
, we have:
![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](//latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
For infinitely many
,
must be an integer, which is not possible.
Therefore, there cannot be infinitely many such
.

Show that for a fixed pair of distinct positive integers



![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)

Let
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
Then,
![\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/1/9/f/19f2627a432354a757d3c9e8ba42e96f83f46726.png)
![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
Therefore,
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
Let
![\[
(n + a)(n + b) = k^2.
\]](http://latex.artofproblemsolving.com/0/e/c/0ec5e7bac2aebe9ec507727ed21f7ca028655222.png)

![\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]](http://latex.artofproblemsolving.com/7/5/f/75fc52b6662df3a4c0cb2f40f49c4b3351c90a4d.png)

Without loss of generality, we take the first case:
![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](http://latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
Thus,
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](http://latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
Since

![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](http://latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
For infinitely many


Therefore, there cannot be infinitely many such

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