Two consecutive whole cubes

That is, (a + 1)^3 and a^3

Let's consider the difference between the two

(a + 1)^3 - a^3

we get

a^3 + 3a^2 + 3a + 1 - a^3

= 3a^2 + 3a + 1

I'm interested in finding the set of numbers of this difference being a square, that is

(a + 1)^3 - a^3 = n^2

3a^2 + 3a + 1 = n^2

For example, (8^3) - (7^3) = 169 = 13^2

Then, we can show that

n = k^2 + (k +1)^2

for some integer k.

We need to determine "a" such that 3a^2 + 3a + 1 = n^2

This is given in the OEIS

http://oeis.org/A001921

0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, 5570039304932024, 77580639439715775, 1080558912851088832, ...

The recurrence for this sequence is a(n) = 14a(n-1) - a(n-2) + 6.

And,

(104+1)^3 - (104^3) = 32761 = 181^2

(1456^3) - (1455^3) = 6355441 = 2521^2

Giving us the sequence A001570

http://oeis.org/A001570

Numbers n such that n^2 is simultaneously square and centered hexagonal.

A centered hexagonal number, or hex number, is

http://en.wikipedia.org/wiki/Centered_hexagonal_number

Since n^2 = 3a^2 + 3a + 1

then 4n^2 = 12a^2 + 12a + 4
And 4n^2 - 1 = 12a^2 + 12a + 3
4n^2 - 1 = 3(2a + 1)^2

If p is prime and p|(2a + 1)^2,

then p|2n + 1 or p|2n − 1,
but p cannot divide both (2n + 1) and (2n - 1)
Otherwise, p|(2n+1)−(2n−1) = 2, which cannot be since (2a+1)^2 is odd.

We now have two possible cases (u and v are integers):
2n + 1 = u^2 and 2n - 1 = 3v^2,
or 2n + 1 = 3u^3 and 2n - 1 = v^2

The former case implies
2n = u^2 - 1 = 3v^2 + 1 or u^2 = 3v^2 + 2 = 2 (mod 3)
which is impossible since for any integer u, u^2 = 0 or 1 (mod 3)
Therefore,
2n - 1 = v^2 for some odd positive integer v.

Let k = (v - 1)/2.
Then
k^2 + (k + 1)^2 = 2k^2 + 2k + 1 = {(2k + 1)^2 + 1}/2 = (v^2 + 1)/2 = n.