If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$ If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$
I attempted to solve this question using binomial coefficients, saying:
$(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}=\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a$
and from here I was thinking that I had to do something with $\sqrt{m}^{n-a}$ with $\sqrt{k}$ and do something else with $\sqrt{k+1}$ and $\sqrt{m+1}^a$. Unfortunately I couldn't think of what to do with this circumstance.
I then thought of maybe taking $(\sqrt{k}+\sqrt{k+1})^2=(\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a)^2$, but this immediately over-complicated the question.
Could you please explain to me how to solve this question and how to solve similar questions in the future?
 A: This is a USAMTS problem. The official solution can be found here. You are actually on the right track.
Below is my solution. Notice that I used the same notation as in the original problem.

My proof:
We prove by induction that: If $m$ is odd then
$$x^m = a_m\sqrt n + b_m \sqrt{n+1}, na_m^2+1 = (n+1) b_m^2 $$
If $m$ is even then
$$x^m = c_m + d_m \sqrt{n(n+1)}, c_m^2 = n(n+1) d_m^2 +1 $$
When $m=1$ it's true.
If it's true for $m$, we prove it's true for $m+1$.
Case 1: If $m$ is odd, then
$$x^{m+1} = (a_m\sqrt n + b_m \sqrt{n+1}) (\sqrt n + \sqrt{n+1})$$
$$=na_m + (n+1) b_m + (a_m+b_m) \sqrt{n(n+1)}$$
So $c_{m+1} = na_m+(n+1)b_m, d_{m+1} = (a_m+b_m)$ and
$$c_{m+1}^2-n(n+1)d_{m+1}^2-1$$
$$= n^2 a_m^2 + 2n(n+1)a_mb_m + (n+1)^2b_m^2 - n(n+1)a_m^2 - 2n(n+1)a_mb_m - n(n+1)b_m^2 - 1$$
$$
= - n a_m^2+(n+1)b_m^2 - 1 = 0
$$
Therefore $x^{m+1}$ is groovy.
Case 2: If $m$ is even then
$$x^{m+1} = (c_m+d_m\sqrt{n(n+1)})(\sqrt n + \sqrt{n+1}) $$
$$=(c_m+(n+1)d_m)\sqrt n + (c_m+nd_m)\sqrt{n+1}$$
So $a_{m+1} = c_m+(n+1)d_m, b_{m+1} = c_m + nd_m$ and
$$na_{m+1}^2+1 - (n+1) b_{m+1}^2$$
$$=nc_m^2+2n(n+1)c_md_m+n(n+1)^2 d_m + 1 - (n+1)c_m^2 - 2n(n+1) c_md_m - (n+1) n^2 d_m^2$$
$$=-c_m^2 + (n+1) d_m^2 + 1=0$$
Again $x^{m+1}$ is groovy.
A: Let $ x = \sqrt{m+1} + \sqrt{m}$ and $ y = \sqrt{m+1} - \sqrt{m}$.
Observe that (IE Fill in the gaps to prove these statements. If you're stuck, explain what you've tried.)

*

*$x \geq 1$

*$xy = 1$

*Since $ f(a) = \sqrt{a+1} + \sqrt{a}$ is strictly increasing with image $[1, \infty)$, so there is a unique real (not necessarily integer) $k_n$ such that $x^n = \sqrt{ k_n+1} + \sqrt{k_n}$. It remains to show that $k_n$ is indeed an integer.

*$\sqrt{ k_n+1} - \sqrt{k_n} = \frac{1}{ \sqrt{k_n+1} + \sqrt{k_n}}  = \frac{1}{x^n} = y^n .$

*$\sqrt{k_n } = \frac{1}{2} (x^n - y^n) $

*$k_n = \frac{1}{4} (x^{2n} + y^{2n} - 2 ) =  \frac{1}{2} \left[  -1 + \sum_{i=0}^n {2n \choose 2i}m^i (m+1)^{n-i}           \right] $.

*Now, show that this is an integer, because the numerator is even.


 Consider seperate cases of $m$, $m+1 $ even.
 Notice that all but one of the terms in the summation are even.

Notes

*

*Having read the official solution, I'm amazed this (natural to me) approach worked out so quickly but they didn't use it.

*Defining $y$ is pretty natural given $x$, and it's a useful trick in such situations. (EG It's also in the official solution.)

*Bonus: $ y^n = \sqrt{ k_n + 1 } -\sqrt{k_n}$

*$k_1 = m, k_2 = 4m^2 + 4m, k_3 = 16m^3 + 24m^2 + 9m$

*I'm slightly amazed that there is a closed form expression for $k_n$. I looked at initial $k_n$ and couldn't spot a generalizable pattern.

