Algebrically prove that the difference of the squares of any two consecutive integers is an odd integer

Let  be the first integer. the next integer must then be  .the difference of their squares is an odd integer, so    is expandable.      therefore    all even  integers  are divisible by 2. forces  to always be an even integer. it forces 2 to be a factor of  .if you add 1 to any even integer, you get an odd integer.so 2n makes always  an even integers ; adding 1 therefore makes it always odd.

Hey, can you post the problem?