Prove that $(2m+1)^2 - 4(2n+1)$ can never be a perfect square where m, n are integers I could prove it hit and trial method. But I was thinking to come up with a general and a more 'mathematically' correct method, but I did not reach anywhere. Thanks a lot for any help.
 A: Suppose to the contrary that the difference is a perfect square. Note that the difference is odd, so we would have
$$(2m+1)^2-4(2n+1)=(2q+1)^2.$$
This can be rewritten as 
$$4(2n+1)=4(m^2+m)-4(q^2+q).\tag{1}$$
Note that both $m^2+m$ and $q^2+q$ are always even, so the right-hand side of (1) is divisible by $8$. But the left-hand side is not, and we have reached a contradiction.  
A: The presence of all those even numbers and the fact you're interested in squares both suggest that you consider the problem modulo powers of $2$.
Usually, $8$ is the go-to modulus for squaring problems, since that's where squaring odd numbers displays its full weirdness (all odd numbers square to $1$ mod $8$) and going to higher powers of $2$ doesn't usually yield more information. But sometimes $4$ is good enough, and some problems would indicate looking at higher powers of $2$ after some analysis (e.g. even numbers are involved).
But in this case, $8$ is good enough:
$$ \mathrm{odd}^2 - 4 \cdot \mathrm{odd} \equiv 1 - 4 \equiv 5 \pmod{8} $$
and no number can square to $5$ modulo $8$.
A: Suppose that $(2m+1)^2-4(2n+1)=a^2$
Since $2m+1$ is odd and $4(2n+1)$ is even $a$ must be odd  so  $$(2m+1)^2-4(2n+1)=(2k+1)^2$$
which gives $4(2n+1)=(2m+1-2k-1)(2m+1+2k+1)\rightarrow4(2n+1)=2(m-k)\cdot2(m+k+1)$
 so, $2n+1=(m-k)\cdot(m+k+1)$
But $(m-k)+(m+k+1)=2m+1$  so $(m-k)$ and $(m+k+1)$ have not the same parity which means that one of them is even and so is their product.
(which as we saw is  just the odd number $2n+1$)
This gives us the desired contradiction.
A: If $(2m+1)^2-4(2n+1)$ is a square, then it has the form $q^2$ where $q$ is an odd number (since $(2m+1)^2-4(2n+1)$  is odd). In this case, the quadratic equation
$$x^2-(2m+1)x+(2n+1)=0$$
has a pair of solutions which are integers (since $q$ is odd).
But both the sum $(2m+1)$ and the product $(2n+1)$ of such solutions are odd. This is a contradiction (since for arbitrary integers $a$ and $b$, at least one of $a+b$ and $ab$ must be even).
A: an odd^2 is always of the form 8k+1.now if it is a ssquare then it wiil be of the form 8k+1.again (2m+1)^2 will be of the form some 8t+1 i.e. 4(2n+1) will be divisible by 8 which leads to contradiction
