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
a=3*k^2+3k+1= (k+1)^3-k^3 qed
Edited on March 25, 2017, 9:10 pm