<s>[INST] 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.
 [/INST] 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
$$4m+1=4(a^2+a)+1=(2a+1)^2. $$
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