Let n and a be positive integers such that 3n^2=4a-1.

Prove that a is a difference of consecutive cubes.

4a-1 is ODD and divisible by 3

We can assume 4a-1= 3*(2*k+1)^2=12*k^2+12*k+3

4*a=4*(3*k^2+3k+1)

a=3*k^2+3k+1= (k+1)^3-k^3 qed

*Edited on ***March 25, 2017, 9:10 pm**