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.
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.
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