Prove that $4m + 1$ is a perfect square if $\{ \sqrt {n + \sqrt n}\} = \{\sqrt m\}$ 
Let $n,m \in \mathbb{N}-\{0\}$ so that $\{ \sqrt {n + \sqrt n}\} =
 \{\sqrt m\} \tag1$ Prove that $4m + 1$ is a perfect square.

($\{x\}$ is the fractional part of $x$)

No idea how to start. Replacing $\{x\}$ by $x-[x]$ doesn't seem to help.
Trying some values for $n, m$, it seems that (1) cannot hold unless $n$ is a perfect square.
 A: So $$\tag1\sqrt{n+\sqrt n}-\sqrt m$$ is an integer $k\in\Bbb Z$.
Then 
$\alpha_=\sqrt{n+\sqrt n}$ is a root of the polynomial
$$f(X)=X^4-2nX^2+n^2-n$$ 
but because   $\alpha=\sqrt m+k$, it is also a root of
$$g(X)=X^2-2kX+k^2-m $$
hence also a root of 
$$\tag2\begin{align}&f(X)-(X^2+2kX+3k^2+m+2n)g(X)\\&=4k(k^2+m-n)X +(-3k^4+2(n+m)k^2+(n-m)^2-n)\end{align}$$
If $4k(k^2+m-n)\ne 0$, this implies that $\sqrt m$ and $\sqrt n$ are rational, so $m,n$ are perfect squares, say $n=a^2$, $m=b^2$ with $a,b>0$, and also $n+\sqrt n=a^2+a$ is a perfect square, say $a^2+a=c^2$ with $c>0$.
But then $a=c^2-a^2=(c+a)(c-a)$, contradicting $|c+a|>|a|>0$.
If $k^2+m-n=0$, the constant term in $(2)$ must be zero, i.e.,
$$\begin{align}0&=-3k^4+2(n+m)k^2+(n-m)^2-n 
\\&= -3(n-m)^2+2(n+m)(n-m)+(n-m)^2-n\\
&=(4m-1)n-4m^2 \end{align}$$
i.e., $4m^2=(4m-1)n$.
As $4$ and $m$ are coprime to $4m-1$ and $4m-1\ge 3$, we arrive at a contradiction.
Therefore, we are left only with the case $k=0$.
But then $n+\sqrt n=m$, so that $\sqrt n$ is rational, hence a perfect square, say $n=a^2$, and finally
$$4m+1=4(a^2+a)+1=(2a+1)^2. $$
A: In this solution $\mathbb{N}$ denotes the set of positive integers.  For the set of nonnegative integers, $\mathbb{N}_0$ is used.
Note that $\mathbb{Q}\big[\sqrt{n+\sqrt{n}}\big]=\mathbb{Q}\big[\sqrt{m}\big]$.  If $\sqrt{m}$ is rational, then $\sqrt{n+\sqrt{n}}$ is also rational.  Thus, $n$ is a perfect square, and $$\frac{1}{2}\sqrt{(2a+1)^2-1}=\sqrt{a^2+a}=\sqrt{n+\sqrt{n}}\in\mathbb{Q}$$
if $n=a^2$ for some $a\in\mathbb{N}_0$. That is $(2a+1)^2-1$ is a perfect square, or $a=0$.  Then, $m=n=0$, which is a contradiction.  Hence, $\sqrt{m}$ is not a rational number.
We now see that $\sqrt{n+\sqrt{n}}\notin\mathbb{Q}$ lies in $\mathbb{Q}\big[\sqrt{m}\big]$.  As a minimal polynomial of $\sqrt{n+\sqrt{n}}$ in $\mathbb{Q}[X]$ divides $X^4-2n\,X^2+\left(n^2-n\right)$, the conjugate in $\mathbb{Q}\big[\sqrt{m}\big]$ of $\sqrt{n+\sqrt{n}}$ is one of the three possible values: (a) $\sqrt{n-\sqrt{n}}$, (b) $-\sqrt{n-\sqrt{n}}$, or (c) $-\sqrt{n+\sqrt{n}}$.
In the first two cases (a) and (b), a minimal polynomial of $\sqrt{n+\sqrt{n}}$ must then take the form $X^2-u\,X+v$ for some $u,v\in\mathbb{Q}$.  Note that $|v|=\sqrt{n^2-n}$, so $\sqrt{n^2-n}=\frac{\sqrt{(2n-1)^2-1}}{2}$ is rational.  This means $n=0$ or $n=1$.  
In the last case (c), a minimal polynomial of $\sqrt{n+\sqrt{n}}$ must take the form $X^2-w$ for some $w\in\mathbb{Q}$.  Note that $w=n+\sqrt{n}$, so $n$ is a perfect square.  That is, $n=a^2$ for some $a\in\mathbb{N}_0$. 
From the assumption that $m,n\neq 0$, the previous two paragraphs yield $n=a^2$ for some $a\in\mathbb{N}$.  Now, we use the given equality $\sqrt{n+\sqrt{n}}=\sqrt{m}+k$ for some $k\in\mathbb{Z}$.  This means $\sqrt{m}+k=\sqrt{a^2+a}$.  Thus, $$m+k^2+2k\sqrt{m}=a^2+a\in\mathbb{Q}\,.$$
Since $\sqrt{m}$ is irrational, we must have $k=0$.  Ergo, $m=a^2+a$, and $4m+1=(2a+1)^2$, as required.
P.S.: The condition $\mathbb{Q}\big[\sqrt{n+\sqrt{n}}\big]=\mathbb{Q}\big[\sqrt{m}\big]$ already implies $n=a^2$ and $m=r^2\left(a^2+a\right)$ for some $a\in\mathbb{N}$ and $r\in\mathbb{Q}_{>0}$.  The given equality only establishes $r=1$.
