To celebrate my birthday, I like to find interesting number theoretic
properties of my new age. My upcoming 61st birthday was challenging, but then
I noticed that $61 = 5^2 + 6^2 = 5^3 - 4^3$, the sum of two consecutive squares
and the difference of two consecutive cubes. I wondered what other numbers
had this property; that is, the integer solutions to
$a^2+(a+1)^{2} = (b+1)^3 - b^3$ or equivalently $2a^2+2a = 3b^2+3b$.
I ran an experiment in Matlab and got the following striking results.


$61 = 5^2 + 6^2 = 5^3 - 4^3$


$5941 = 54^2 + 55^2 = 45^3 - 44^3$


$582121 = 539^2 + 540^2 = 441^3 - 440^3$


$57041881 = 5340^2 + 5341^2 = 4361^3 - 4360^3$


$5589522181 = 52865^2 + 52866^2 = 43165^3 - 43164^3$


$547716131821 = 523314^2+523315^2 = 427285^3-427284^3$


$53670591396241 = 5180279^2+5180280^2 = 4229681^3 - 4229680^3$


It is easy to show that $a/b$ is close to $\sqrt{3/2}$ but not very close; the
error is $O(1/b)$. The ratio between successive values of
$a$ is monotonically decreasing; its last value is 9.898987988091280
so perhaps it is converging to 980/99.


Does anyone know what is going on here?



Sum of two consecutive squares equals difference of two consecutive cubes
There is the following sequence of positive solutions (according to Mathematica): $a\_n=\frac{1}{8} \left(\left(\sqrt{6}-2\right) \left(2 \sqrt{6}+5\right)^n-\left(\sqrt{6}+2\right) \left(5-2 \sqrt{6}\right)^n-4\right),$ and 
$b\_n=\frac{1}{12} \left(-\left(\sqrt{6}-3\right) \left(2 \sqrt{6}+5\right)^n+\left(\sqrt{6}+3\right) \left(5-2 \sqrt{6}\right)^n-6\right)$.


Your Diophantine equation is known as the [Pell equation](http://mathworld.wolfram.com/PellEquation.html).


