If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$

Question

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?

Answer 1

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.)

1. $x \geq 1$
2. $xy = 1$
3. 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.
4. $\sqrt{ k_n+1} - \sqrt{k_n} = \frac{1}{ \sqrt{k_n+1} + \sqrt{k_n}} = \frac{1}{x^n} = y^n .$
5. $\sqrt{k_n } = \frac{1}{2} (x^n - y^n) $
6. $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] $.
7. 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.

Answer 2

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.

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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.