If $d\mid2n^2$, could $n^2+d$ be a square of a natural number? Here $d, n \in \Bbb N$. If $d\mid2n^2$, could $n^2+d$ be a square of a natural number? Here $d, n \in \Bbb N$.
What I tried so far: $2n^2 = kd, k \in N$ $$d = 2\frac{n^2}{k}$$
$$n^2 + d = $$
$$n^2 +2\frac{n^2}{k}= n^2\left(\frac{k+2}{k}\right) = \frac{kd}{2}\left(\frac{k+2}{k}\right) = \frac{d(k+2)}{2}$$
I don't see where this could lead me. 
 A: Your reasoning looks good. You have that $$n^2+d=n^2\left(\frac{k+2}{k}\right)$$
That is a perfect square only if $\frac{k+2}{k}$ is the square of a rational number. However, a rational square, in lowest terms, is a ratio of two integer squares. No two integers squares differ by $2$, and a possible reduction of $\frac{k+2}{k}$ wouldn't help, because no two integer squares differ by $1$. (Obviously, $0$ and $1$ aren't helpful here.)
We conclude that $n^2+d$ is not the square of an integer.
A: I am assuming, for this answer, that $0\not\in\mathbb{N}$.
The Reasoning in the Question
The question asks whether there are $m,n,d\in\mathbb{N}$ so that $d\mid2n^2$, and $n^2+d=m^2$. This is equivalent to asking if there are $m,n\in\mathbb{N}$ so that $m\gt n$ and $m^2-n^2\mid2n^2$.
We can assume that $(m,n)=1$. If not, then divide $m$ and $n$ by $(m,n)$.
Suppose that
$$
m^2-n^2\mid2n^2\tag1
$$
This means there is a $k$ so that
$$
k\left(m^2-n^2\right)=2n^2\tag2
$$
which means that
$$
(k+2)n^2=km^2\tag3
$$
Finishing with Bezout
Since $(m,n)=1$, this answer says we also have $\left(m^2,n^2\right)=1$. Bezout gives us $a,b$ so that $am^2+bn^2=1$. Therefore,
$$
\begin{align}
k
&=akm^2+bkn^2\tag{4a}\\
&=a(k+2)n^2+bkn^2\tag{4b}\\
&=(ak+2a+bk)n^2\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a):}$ multiply $am^2+bn^2=1$ by $k$
$\text{(4b):}$ apply $(3)$
$\text{(4c):}$ collect multiples of $n^2$
$(4)$ says that $n^2\mid k$. Then, dividing $(2)$ by $n^2$ gives
$$
\frac{k}{n^2}\left(m^2-n^2\right)=2\tag5
$$
That is, $m^2-n^2\mid2$. This means that either $m^2-n^2=2$ or $m^2-n^2=1$.
If $2=m^2-n^2=(m+n)(m-n)$, we must have $m+n=2$ and $m-n=1$ which means $m=\frac32$ and $n=\frac12$.
If $1=m^2-n^2=(m+n)(m-n)$, we must have $m+n=1$ and $m-n=1$ which means $m=1$ and $n=0$.
Since neither $\frac12$ nor $0$ are in $\mathbb{N}$, we cannot have $m,n\in\mathbb{N}$ and $m^2-n^2\mid2n^2$.

A Comment on the Other Answer
While the claim made in G Tony Jacobs' answer, that if $\frac ab=\left(\frac pq\right)^2$ and $(a,b)=1$ then $a$ and $b$ are squares of integers, is true, I think it requires some justification.
Suppose that $(a,b)=1$ and $(p,q)=1$ and
$$
\frac ab=\left(\frac pq\right)^2\tag6
$$
Bezout gives us $x,y$ such that
$$
ax+by=1\tag7
$$
We have
$$
\begin{align}
p^2
&=ap^2x+bp^2y\tag{8a}\\
&=ap^2x+aq^2y\tag{8b}\\
&=a\left(p^2x+q^2y\right)\tag{8c}
\end{align}
$$
Explanation:
$\text{(8a):}$ multiply $(7)$ by $p^2$
$\text{(8b):}$ apply $(6)$
$\text{(8c):}$ factor out $a$
Similarly, we have
$$
\begin{align}
q^2
&=aq^2x+bq^2y\tag{9a}\\
&=bp^2x+bq^2y\tag{9b}\\
&=b\left(p^2x+q^2y\right)\tag{9c}
\end{align}
$$
As mentioned above $(p,q)=1\implies\left(p^2,q^2\right)=1$. Thus, since $(8)$ and $(9)$ say that $p^2x+q^2y$ is a common factor of $p^2$ and $q^2$, we must have
$$
p^2x+q^2y=1\tag{10}
$$
Then $(8)$, $(9)$, and $(10)$ say that $a=p^2$ and $b=q^2$.
