Mordellbles

The $n^\text{th}$ diamond pattern uses $n^2+(n-1)^2$ pebbles, as below:

So the equation we want to solve in positive integers is $m^3=2n^2-2n+1$

Multiply by $8$ : $8m^3=16n^2-16n+8=(4n-2)^2+4$

Put $x=2m$ and $y=4n-2$ to get $y^2=x^3-4$

which is an example of Mordell's equation . Better still, this precise example is solved on page 6 of this awesome paper by Keith Conrad. He shows the only solutions in integers to the above equation are $(x,y)=(2,\pm2)$ and $(x,y)=(5,\pm11)$ .

The first of these leads to solutions in the original equation $(m,n)=(1,0)$ and $(m,n)=(1,1)$ (but we can't have $n=0$ ). The second gives non-integer values for $(m,n)$ so is discounted.

Hence there are no other solutions .

@Chris Lewis , @David Vreken, @Pi Han Goh: Please, correct me if I am wrong:

Let's $a_{k + 1}$ to be the general term of this series, then

(1) $a_{k + 1} = 1 + \frac{k\times (4k + 4)}{2} = 1 + k\times(2k +2) = 1 + 2k^{2} + 2k$

Now suppose that $a_{k + 1} = m^{3}$ , then

(2) $m^{3} = 1 + 2k + 2k^{2}$

Now I will add on both sides of the equation $k + k^{2} + k^{3}$ , then

(3) $m^{3} + k + k^{2} + k^{3} = 1 + 3k + 3k^{2} + k^{3} = (k + 1)^{3}$ ,

which means that

(4) $k + 1 > k \geq m$

Now, we know that

(5) $a + b + c \geq 3\times (abc)^{\frac{1}{3}}$ so from (1) and (5)

(6) $m^{3} = 1 + 2k + 2k^{2} \geq 3\times 4^{\frac{1}{3}} \times k$

which is a contradiction with (4). Therefore, we can get only a perfect cube with 1 pebble