Given $\{a_n\}$ a sequence of real numbers such that $a_1 > 1$ and $a_{n+1}-a_n\geq 2$, with $S_n=a_1+a_2+...+a_n$, prove that for any $n\geq 3$, there is at least one perfect square in the range $[S_n, S_{n+1}]$.
I was thinking that $a_{n+1}\ge 2+a_n \ge 4+a_{n-1}\geq ... \ge 2n+a_1>2n+1$. And this implies $S_{n+1}-S_n=a_{n+1}>2n+1=(n+1)^2-n^2$. So at least one of $n^2$ or $(n+1)^2$ si between $S_{n}$ and $S_{n+1}$.
Is this reasoning correct? I don't think it is. How to prove the problem then?
On
For the sake of contradiction assume that there exists some $n$ such that $[S_n, S_{n+1}]$ does not contain a perfect square. That means there is some positive integer $p$ such that:
$$p^2 < S_n<S_{n+1} < (p+1)^2$$
We then have:
$$a_{n+1} = S_{n+1}-S_n<(p+1)^2-p^2=2p+1$$
That implies $a_n \leq a_{n+1}-2<2p-1$ and generally $a_{n-k} < 2p-(2k+1)$ for any $k\leq n-1$. Summing up all these inequalities we get:
$$S_n=a_1+a_2+\ldots+a_n<\sum_{k=0}^{n-1}\left[2p-(2k+1)\right]=2np-n^2$$
Therefore:
$$2np-n^2>S_n>p^2\Rightarrow (p-n)^2 < 0$$
which is obviously a contradiction.
Let $(a_n)$ be a sequence of real numbers with $a_1 \geq 0$ and $a_{n+1} - a_n \geq 2$. Define $S_n = a_1 + \cdots + a_n$. We will show that $[S_n, S_{n+1}]$ contains a perfect square for all $n \geq 1$.
Lemma. Let $0 \leq x < y$ be real numbers with $y -x \geq 2\sqrt x + 1$. Then $[x, y]$ contains a perfect square.
Proof. Let $n^2 \geq 0$ be the largest square with $n^2 \leq x$. Then $$x < (n+1)^2 = n^2 + 2n + 1 \leq x + 2 \sqrt x + 1 \leq y \,. \qquad\square$$
So suffices to show that $a_{n+1} \geq 2 \sqrt{S_n} + 1$. I.e. that $4 S_n \leq (a_{n+1} - 1)^2$. We prove this by induction.
For $n = 1$ this follows from $$(a_2-1)^2 \geq (a_1+1)^2 \geq (2\sqrt{a_1})^2 = 4 a_1 \,.$$ Take $n \geq 1$, and suppose $$4 S_n \leq (a_{n+1} - 1)^2 \,.$$ Add $4a_{n+1}$ to both sides to obtain $$4S_{n+1} \leq (a_{n+1} + 1)^2 \,.$$ Then use that $a_{n+1} +1 \leq a_{n+2} - 1$ to get $$4S_{n+1} \leq (a_{n+2} - 1)^2 \,.$$ The induction is complete.