Question 1191208
<pre>

2n-1 is an odd integer for any integer n
The next larger odd integer is 2 more than that or 
2n-1+2 or 2n+1

The sum of their cubes is

(2n-1)<sup>3</sup>+(2n+1)<sup>3</sup>

Factor this as the sum of two cubes:

[(2n-1)+(2n+1)][(2n-1)<sup>2</sup>-(2n-1)(2n+1)+(2n+1)<sup>2</sup>]

[2n-1+2n+1][(2n-1)<sup>2</sup>-(2n-1)(2n+1)+(2n+1)<sup>2</sup>]

[4n][(2n-1)<sup>2</sup>-(2n-1)(2n+1)+(2n+1)<sup>2</sup>]

There is no need to simplify the factor on the right
because the factor on the left is divisible by 4, so
the whole thing is divisible by 4.

Edwin</pre>