# Infinitely many positive integers $m$

Let $a_n = \left[\sqrt{(n+1)^2+n^2}\right], n = 1,2,\ldots,$ where $[x]$ denotes the integer part of $x$. Prove that

$\quad$ (a) there are infinitely many positive integers $m$ such that $a_{m+1}-a_{m} > 1$;

$\quad$ (b) there are infinitely many positive integers $m$ such that $a_{m+1}-a_m = 1$.

If $a_{m+1} \geq a_m+1$, then if $k^2 \leq 2m^2+2m+1 < (k+1)^2 \implies k < 2m$, then $$(k+1)^2 \leq 4m^2.$$

How can we continue this approach or is there another way (I saw another way using Pell equations)?

A famous fact about binomial coefficients is: $$\binom{n}{2}\text{ is a perfect square for infinitely many } n.$$ This statement can be proved without Pell's equation, in the following way: $$\binom{(2n-1)^2}{2}=(2n-1)^2\cdot2n(n-1)$$ and if $\binom{n}{2}$ is a perfect square, the above expression is another perfect square. Now note that $\binom{9}{2}=6^2$.

Thus we may take $m=n-2$ where $\binom{n}{2}$ is a perfect square and finish the proof of (a).
(b) is trivial.

• How does this prove part (a)? – user19405892 Jul 29 '16 at 2:11
• @user19405892 Suppose $n(n-1)=2k^2$. One may immediately check that $a_{n-1}=2k$ and $a_{n-2}\leq 2k-2$(This inequality is done by a plain estimation $n>k+1$). – Cave Johnson Jul 29 '16 at 2:16
• Can you explain how to get $a_{n-2} \leq 2k-2$? How do you know $n > k+1$? – user19405892 Jul 29 '16 at 2:23
• If $n\leq k+1$ then $n(n-1)\leq k(k+1)<2k^2$, a contrdiction. So $\sqrt{(n-1)^2+(n-2)^2}=\sqrt{2n(n-1)-4n+5}<\sqrt{4k^2-4(k+1)+5}=2k-1$. – Cave Johnson Jul 29 '16 at 2:31
• How do we prove (b)? – user19405892 Jul 29 '16 at 2:35

Here is a proof sketch. First we show that $a_{m+1} - a_m$ can never be zero. This is because $(n+1)^2 + n^2$ increases by $4n + 4$ when $n$ increases by $1$. But, consecutive squares near $(n+1)^2 + n^2$ are closer together than this. (They are roughly $2 \sqrt{2} n$ apart). So $\sqrt{(n+2)^2 + (n+1)^2}$ and $\sqrt{(n+1)^2 + n^2}$ differ by more than one, and their floors cannot be the same.

Next, assume (a) is false. There are only finitely many $m$ such that $a_{m+1} - a_m > 1$, so let $n$ be the largest one. For any $m > n$, $a_{m+1} = a_m + 1$. This means, for instance, $a_{2n} - a_n = n$. But the difference between $\sqrt{(n+1)^2 + n^2}$ and $\sqrt{(2n + 1)^2 + (2n)^2}$ is roughly $n\sqrt{2}$; it is too big for the difference of the floors to be $n$.

Similarly, assume (b) is false. Now we can say for any $m > n$, $a_{m+1} \ge a_m + 2$. So $a_{2n} - a_n \ge 2n$. The difference of floors of square roots cannot be this large when the difference of square roots directly is, as above, roughly $n\sqrt{2}$.