Show not possible to find positive whole numbers $m,n$ such that $m^2 − n^2 = 6$. Show not possible to find positive whole numbers $m,n$ such that $m^2 − n^2 = 6$. 
$m^2 − n^2 = 6\implies (m+n)(m-n) = 6; \ m,n$ can be either even or odd. 
If $m$ is odd, $m=2k+1, k\ge 1$; else $m=2k$.  
Similarly, if $n$ is odd, $n=2l+1, l\ge 1$; else $n=2l$. 
There are $4$ possible cases: 
(i) $m,n$ both are odd: $(m+n)(m-n)= 4(k+l+1)(k-l)$, 
(ii) $m,n$ both are even: $(m+n)(m-n)= 4(k+l)(k-l)$, 
(iii) $m,n$ one is odd, other is even: $(m+n)(m-n)= (2(k+l)+1)(2(k-l)\pm 1)$$= 4(k+l)(k-l)+2(k-l)\pm(2(k+l)+1)$$= 4(k+l)(k-l) +4k+1, \ \ 4(k+l)(k-l) -4l - 1$ 
Not possible to pursue further except by proof by contradiction. 

If try by contradiction based proof, need prove that the premise $m^2-n^2=6$ is false: 
(i) $(m+n)(m-n)= 4(k+l+1)(k-l)$$\implies 6 = 4(k^2 + kl + k - kl -l^2 -l)= 4(k^2 -l^2 +k -l) = 4(k-l)(k+l+1)$$\implies 3 = 2(k+l+1)(k-l)$ 
The lhs is odd, but the rhs is even. 
(ii) $(m+n)(m-n)= 4(k+l)(k-l)$$\implies 6 = 4(k+l)(k-l)$
$\implies 3 = 2(k+l)(k-l)$ 
The lhs is odd, but the rhs is even. 
(iii) $4(k+l)(k-l) +4k+1$$\implies 5 = 4(k+l)(k-l)+k)$ 
The lhs is odd, but the rhs is even. 
(iv) $4(k+l)(k-l) -4l - 1$$\implies 7 = 4(k+l)(k-l)-l)$ 
The lhs is odd, but the rhs is even.

However, the same cannot be said for $m^2 - n^2 = 6j, \ m,n,j \ \in \mathbb{N}$.
 A: Another alternative here: $m+n$ and $m-n$ are either both odd or both even i.e. their product is either odd or divisible by $4$ and $6$ is neither.
A: There's an easier way to do this. So, you know $(m+n)(m-n) = 6$, and $6 = 2 \cdot 3 = 1 \cdot 6$. If $m,n$ are to be whole numbers, then so must $m+n$ and $m-n$. Thus, we could look at the systems of equations that result by setting factors equal to each other:
$$\left\{\begin{matrix}
m+n = 2\\ 
m-n=3
\end{matrix}\right. \;\;\;\;\; \left\{\begin{matrix}
m+n = 3\\ 
m-n=2
\end{matrix}\right. \;\;\;\;\; \left\{\begin{matrix}
m+n = 1\\ 
m-n=6
\end{matrix}\right. \;\;\;\;\; \left\{\begin{matrix}
m+n = 6\\ 
m-n=1
\end{matrix}\right.$$
Show that, in none of these cases, you do not have both $m,n$ as whole numbers to get a solution.
A: Note that
$$m^2 =\sum_{k=1}^m(2k-1)\quad \text{and}\quad n^2 =\sum_{k=1}^n(2k-1)$$
Hence,
$$m^2-n^2 = \sum_{k=n+1}^m(2k-1)\stackrel{!}{=}6$$
It follows that $m\leq 3$ and $n+1\geq 2$, but the possible sums are only $3,5,3+5$. So, there is no solution.
A: Seems fine though your proof is long. 
As for your question regarding generalization to $6j$. 
For the special case where $j$ is odd, we also can't have $m^2-n^2=6j$.
Suppose not, then we have $(m+n)(m-n)=6j$. 
If $m+n$ and $m-n$ are both odd, then the LHS is odd but RHS is even.
If $m+n$ and $m-n$ are both even, then LHS is a multiple of $4$, but RHS is not a multiple of $4$.
A: The answers already added are pretty good in my opinion, especially the one by @trancelocation. However, since nobody has approached this via congruences yet, I wanted to do the same.
We consider the general case, $m^2-n^2 = 6j, m,n,j \ge 1$. It is an elementary fact that any square is congruent to $0,1$ modulo $4$ according to as it is even or odd. Then, rewriting the equation, we have,
$m^2 = n^2 + 6j \equiv n^2 + 2j \ \bmod 4$
If $n$ is even, we must have $2j \equiv 0 \ \bmod 4 \implies j \equiv 0 \ \bmod 2$.
If $n$ is odd, we again have $2j +1\equiv 1\ \bmod 4 \implies j \equiv 0 \ \bmod 2$.
Hence, the general equation only has solutions if $j$ is even.
