# How to prove that for $a_{n+1}=\frac{a_n}{n} + \frac{n}{a_n}$ , we have $\lfloor a_n^2 \rfloor = n$?

Let $$(a_n)_{n\ge 1}$$ be the sequence defined as the following : $$a_1=1 ,\ a_{n+1}=\dfrac{a_n}{n} + \dfrac{n}{a_n} ,\ n\ge1$$ Show that for every $$n\ge4,\ \lfloor a_n^2 \rfloor = n$$.
My approach to this problem was trying induction and using the function $$f_n(x)=\dfrac{x}{n} + \dfrac{n}{x}$$ :
Proving the base case for $$n= 4$$ and then by the inductive hypothesis $$\lfloor a_n^2 \rfloor = n$$ implies that $$\sqrt{n} \le a_n \lt \sqrt{n+1}$$
We then apply $$f_n$$ knowing that it is decreasing in that interval following it up with the floor function and some polishing, all leads to this inequality : $$n+1\le \lfloor a_{n+1}^2 \rfloor \le n+2$$
So I can't exactly get $$n+1$$ since $$n+2$$ is a possibility, this problem is a product of the fact that if $$a\lt b$$ then $$\lfloor a \rfloor \le \lfloor b \rfloor$$.

Any insights would be greatly appreciated! I wonder if my result is correct because it seems like the only way.

We will prove the following result instead.

$$n+{2\over n}<{a_n}^2

First we check it's true for $$n=4$$.

Now the induction step, first notice that $$x>y>1$$ implies $$x+{1\over x}>y+{1\over y} \equiv 1>{1\over xy}$$

Therefore for RHS $${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{n+{2\over n}\over n^2}+{n^2 \over n+{2\over n}}+2=2+{n^4+n^2+4+{4\over n^2}\over n^3+2n}$$$$= 2+n-{n^2-4-{4\over n^2}\over n^3+2n}

For LHS,

$${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2>{n+1\over n^2}+{n^2\over n+1}+2={n^4+n^2+2n+1\over n^3+n^2}+2$$ $$=2+(n-1)+{2n^2+2n+1\over n^3+n^2}>n+1+{2\over n}>n+1+{2\over n+1}$$

• Nice derivation! although this doesn't really answer the question because $n+\dfrac{2}{n} \lt a_n^2 \lt n+1$ only implies $n\le \lfloor a_{n}^2 \rfloor \le n+1$ from which we can't deduce $\lfloor a_n^2 \rfloor = n$ if I'm not mistaken Aug 14, 2020 at 13:36
• $n+\dfrac{2}{n} \lt a_n^2 \lt n+1$ implies ${a_n}^2$ is strictly less than $n+1$. There is no equality involved here. Aug 14, 2020 at 13:38
• As shown in the $RHS$ part, $2+n-{n^2-4-{4\over n^2}\over n^3+2n}$ is strictly less than $n+2$ as well. Aug 14, 2020 at 13:40
• Ohhh you are right because $n+1$ is an integer if the RHS wasn't an integer we must then have the equality case right ? Aug 14, 2020 at 13:41
• Thanks a lot that is really what I was missing! Aug 14, 2020 at 13:42