Proving that there is a perfect square between $n$ and $2n$ Problem: Prove that there is at least one perfect square in the sequence $n, n+1,\ldots, 2n$.
I know this is in fact very easy, but I can't seem to put my finger on the right equation for it. It involves something like $2(n+1)^2 <(2n)^2$ or the like. 
 A: Suppose that $n\gt x^2$ and $(x+1)^2\gt 2n$ so that there is no square between $n$ and $2n$, then $$(x+1)^2\gt 2n\gt2x^2$$ Because you are dealing with integers and the inequalities are strict you get $$(x+1)^2\ge 2x^2+2$$ which simplifies to $$0\ge x^2-2x+1=(x-1)^2$$
The only value for which this is possible is $x=1$, but you can eliminate that easily enough.
Note: You can use $(x-1)^2+1$ with more care, since one step involves doubling a strict inequality between integers, and then the conclusion is immediate.
A: For all $n$$$ n \leq \lceil \sqrt{n} \rceil^2 \leq 2n.$$
One inequality is trivial, and the other takes some work. 
EDIT:
Using the trivial bound $\lceil \sqrt{n} \rceil \leq \sqrt{n}+1$ we get
$$\lceil \sqrt{n} \rceil^2 \leq n+2\sqrt{n}+1 $$
and we wish this to be lesser than or equal to $2n$. In other words, if $2 \sqrt{n}+1 \leq n$ we're good. Substituting $\sqrt{n}:=x$ gives the quadratic inequality $x^2-2x-1 \geq 0$, which holds for $x \geq 1/2+\sqrt{3}/2$. Thus the only case which we're left with is $n=1$, which obviously holds as well.
A: For $x\geq 6$, there is always an integer in the interval
$$
[\sqrt x,\sqrt{2x}]
$$
which is not hard to prove considering the interval length.

Note that the interval length $f(x)=\sqrt{2x}-\sqrt x=\sqrt x(\sqrt 2-1)$ is strictly growing and
$$
f(x)=1\iff\sqrt x=\sqrt 2+1
$$
so for $x\geq (\sqrt 2+1)^2=3+2\sqrt 2\approx 5.82$ the interval is longer than $1$.
A: The greatest possible square less than $n$ is $\left\lfloor\sqrt{n}\right\rfloor^2$ then we are to examine whether $\left(\left\lfloor\sqrt{n}\right\rfloor+1\right)^2$ satisfies, $$n\le \left(\left\lfloor\sqrt{n}\right\rfloor+1\right)^2\le 2n$$
Proving that $n\le \left(\left\lfloor\sqrt{n}\right\rfloor+1\right)^2$ is trivial. and so we  need to show that $\left\lfloor\sqrt{n}\right\rfloor+1\le\left\lfloor\sqrt{2n}\right\rfloor$. This is also trivial since, $$\sqrt{2n}-1\le \left\lfloor\sqrt{2n}\right\rfloor$$and, $$\left\lfloor\sqrt{n}\right\rfloor\le \sqrt{n}+1$$ Therefore $$\left\lfloor\sqrt{2n}\right\rfloor-\left\lfloor\sqrt{n}\right\rfloor\ge \sqrt{n}\left(\sqrt{n}-1\right)-2$$From which, I hope you can handle the rest.
A: Yet another solution:
First of all, verify that the statement is true for all $n \leq 9$. Then, show by induction that $2k+1 \leq k^2$ for all $k\geq 3$.
Now assume $n > 9$. Let $k$ be the greatest number with $k^2 < n$. Then, $k\geq 3$. Altogether, we get
$$n \leq (k+1)^2=k^2+2k+1 \leq 2k^2 < 2n$$
So, $(k+1)^2$ lies between $n$ and $2n$.
