# 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.

#### Solutions Collecting From Web of "Proving that there is a perfect square between $n$ and $2n$"

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. 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. 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. 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$1$. 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\$.