Proof by induction: one of $n, n+1,\ldots,2n$ is a square 
Show that for every positive integer $n$, one of the numbers $n, n +
1, n + 2, \cdots , 2n$ is the square of an integer.

here what i did. but I am doing some thing wrong here and iam not sure what.
Base Case: $n = 1$ 
$1, 2$ is the square of an integer, which is true since $1$ is the square of an integer.
induction step:
$(k+1)^2 \le 2k^2 + 2$
$k^2 = 2k + 1 \le 2k^2 + 2$
$k^2 - 2k + 1 \ge 0$
$(k-1)^2 \ge 0$
which is true since $(k-1)^2$ is a square, so must be nonnegative.
 A: You have proven the following:

For any $x \in \Bbb{R}$, $(x+1)^2 \le 2x^2+2$.

This will come in handy later.
Inductive Step: Assume that one of the numbers $n,n+1,n+2,...,2n$ is a perfect square, where $n \in \{1,...,k\}$. In particular (for $n=k$), we will assume that for some $r\in \{0,...,k\}$, $k+r$ is a perfect square.
It remains to prove our claim true for $n=k+1$. That is, we must prove that one of the numbers $k+1,k+2,k+3,...,2k+2$ is a perfect square. Now by the induction hypothesis, we know that $k+r$ is a perfect square for some $r\in \{0,...,k\}$. Hence, if $r\in \{1,...,k\}$, then we are done.
It remains to prove the claim true for $r=0$. Since $k+0$ is a perfect square, we have $k=a^2$ for some positive integer $a$. The next perfect square must therefore be at $(a+1)^2$. This is where your proof comes into play. Let $x=a$. Then:
$$
k=a^2<(a+1)^2\le2a^2+2=2k+2
$$
So for the case when $r=0$, one of the integers from $k$ to $2k+2$ must be the perfect square $(a+1)^2$, as desired. This completes the induction.
A: I don't understand your argument after the line "induction step:". But maybe you meant the right thing.
We have to prove by induction that each set $$A_n:=\{n,n+1,\ldots, 2n\}\qquad(n\geq1)$$ contains a square.
The set $A_1=\{1,2\}$ contains a square.
Assume that $n\geq1$ and that $A_n$ contains a square. Then
$$A_{n+1}=A_n\cup\{2n+1,2n+2\}\setminus\{n\}$$
obviously contains a square, unless the only square in $A_n$ was the number $n=k^2$. In this case
$$n<(k+1)^2=k^2+2k+1=n+\sqrt{4n}+1\leq n+\sqrt{4n+(n-1)^2}+1= 2n+2\ ,$$
which proves that $(k+1)^2\in A_{n+1}$.
