# Find all possible integers $n$ such that $\sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.

Find all possible integers $$n$$ such that $$m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$$ is an integer.

Guess what? This problem is adapted from a recent competition. There have been a solution below for you to check out. I am aware of the fact that there are other solutions that are more practical and suitable in test setting.

We have that $$m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}} \ (m \in \mathbb N)$$

$$\iff m - \sqrt{n + 2} = \sqrt{n + \sqrt{n + 2}} \iff (m - \sqrt{n + 2})^2 = n + \sqrt{n + 2}$$

$$\iff m^2 - (2m - \sqrt{n + 2})\sqrt{n + 2} = (\sqrt{n + 2} + 1)\sqrt{n + 2} - 2$$

$$\iff m^2 + 2 = (2m + 1)\sqrt{n + 2} \iff \sqrt{n + 2} = \frac{m^2 + 2}{2m + 1}$$

As an addition, $$\dfrac{m^2 + 2}{2m + 1} \in \mathbb Q^+, \forall m \in \mathbb N \implies \sqrt{n + 2} \in \mathbb Q^+$$

$$\implies \sqrt{n + 2} \in \mathbb N \implies \dfrac{m^2 + 2}{2m + 1} \in \mathbb N \iff \dfrac{4(m^2 + 2) - (2m + 1)(2m - 1)}{2m + 1} \in \mathbb N$$

$$\iff \dfrac{9}{2m + 1} \in \mathbb N \iff 2m + 1\mid 9 \iff 2m + 1 \in \{1, 3, 9\} \iff m \in \{0, 1, 4\}$$

We can set up a table for different value of $$m$$ and $$\sqrt{n + 2}$$.

$$\begin{matrix} m&& 0&& 1&& 4\\ \sqrt{n + 2} = \dfrac{m^2 + 2}{2m + 1}&& 2&& 1&& 2 \end{matrix}$$

$$\iff n \in \{-1, 2\}$$.

Plugging $$n \in \{-1, 2\}$$ in $$m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$$, we have that $$(m,n) = (1, -1)$$ and $$(m, n) = (4, 2)$$ is the correct answer.

• Also, $n=-1$ is valid. Jun 13, 2019 at 10:42

We have that $$n+2 = a^2, n + a = b^2$$, where $$a$$ and $$b$$ are non-negative integers.

If $$a > 2$$, then $$a^2 = n+2 < n+a = a^2 + a - 2 < a^2 + 2a + 1.$$

This means that $$n+a$$ cannot be a perfect square as it's bounded between 2 consecutive perfect squares, which is a contradiction. Hence $$0 \leq a \leq 2$$.

If $$a= 0$$, then $$n = -2$$ but $$n+a = -2$$ is not a perfect square. No solution.
If $$a = 1$$, then $$n = - 1$$ and $$n+a = 0$$ is a pefect square. This gives $$m=1$$.
If $$a=2$$, then $$n=2$$ and $$n+a = 4$$ is a perfect square. This gives $$m=4$$.

A shorter approach:

Generally for numbers $$m=\sqrt{n+a}+\sqrt{n+\sqrt{n+a}}$$ we must have:

$$n+a=b^2$$

$$b^2+b-a=c^2$$

If this system of equations, with given a, has integer solutions for n then m is integer.For example $$a=3$$; $$m=\sqrt{n+3}+\sqrt{n+\sqrt{n+3}}$$ we have:

$$n+3=b^2$$

$$b^2+b-3=c^2$$

It can be seen that with $$c=3$$ we have:

$$b^2+b-3=3^2=9$$; or $$b^2+b-12=0$$ which gives $$b=3$$ and $$b=-4$$. With $$b=3$$ we get $$n=6$$ and $$m=6$$, with $$b=-4$$ we get $$n+3=16$$ or $$n=13$$ which gives $$m=4+\sqrt{17}$$ which is not integer.

In question $$a=2$$ and we can search and see that reasonable value of c can only be zero. Other values give irrational numbers for one of terms $$\sqrt{n+2}$$ or $$\sqrt{n+\sqrt{n+2}}$$. Hence we have:

$$b^2+b-2=0$$

Which gives:

$$b=1$$; $$a= 1$$; $$n=-1$$; $$m= 1$$

$$b=-2$$;$$a= 4$$; $$n=2$$; $$m=4$$

• There are several mistakes. With $n+2=a$, then $\sqrt{n + \sqrt{n + 2}} = \sqrt{a- 2 +\sqrt{a}}$, but you have it being $\sqrt{a+2 +\sqrt a -2} = \sqrt{a +\sqrt{a}}$ instead. However, your later formula of $b^2 + b - 2$ handles this correctly. Nonetheless, you don't explain why you can assume that $b^2 + b - 2 = 0$. Finally, with your last $2$ lines, if $b = 1$, then $a = 1$, not $a = \pm 1$, and if $b = -2$, then $a = 4$, not $a = \pm 2$. Jun 13, 2019 at 21:58
• @JohnOmielan, thanks for correcting comment. I edited my answer and explained why second term is equal to zero. Jun 14, 2019 at 7:29
• Your welcome. However, you still haven't corrected my first stated mistake of not correctly stating $\sqrt{a - 2 + \sqrt{a}}$. Also, once again, if $b = -2$, then $a = 4$ (and, thus, $n = 2$), but you have it as $a = 2$. Jun 14, 2019 at 8:20