Consecutive cubes equal to a square $\frac{1}{8}ab(a^2+b^2-1) = y^2$, and Pythagorean triples If we wish that the sum of $b$ consecutive cubes with initial cube $c_k=\tfrac{1}{2}(1+a-b)$ is equal to a square, then we have the rather simple equation,
$$F_k=\tfrac{1}{8}ab(a^2+b^2-1) = y^2$$
It seems only three integral families are known that solve it:
$$\begin{align}
F_1(m):\quad a,\,b &= m+1,\;\; m\\
F_2(n):\quad a,\,b &= 16n^3+24n^2+8n,\;\; 2n+1\\
F_3(n):\quad a,\,b &= 4n^4-1,\;\; 2n^2\\
\end{align}$$
Examples.


*

*$F_1(22)$ and $F_1(27)$


$$1^3+2^3+\dots+22^3 = 253^2$$
$$1^3+2^3+\dots+27^3 = 378^2$$


*$F_2(1)$ 


$$23^3+24^3+25^3 = 204^2$$


*$F_3(2)$: 


$$28^3+29^3+\dots+35^3 = 504^2$$
This implies that $F_1(c_2-1)+F_2(n)$ or $F_1(c_3-1)+F_3(n)$ are in a Pythagorean triple,
$$F_1(22)+F_2(1) = 253^2+204^2 =325^2$$
$$F_1(27)+F_3(2) = 378^2+504^2 =630^2$$

Question: Can anyone give a fourth integral family $F_4$ where initial cube $c\neq0,1$? 

 A: There exists an entire family of families by choosing $a=kb$ ($b$ odd, $k$ even and appropriate):
It is then sufficient to solve $(k^2+1)b^2-1=2kx^2$. This is just an almost Pell-type equation and hence for fixed $k$ has infinitely many solutions which can easily be parametrised, provided at least one solution exists and if $k$ is chosen so that $2k(k^2+1)$ is not a square (but that is impossible for $k>1$ even).
Many such $k$ exist, for example every $k=2m^2$ has initial solution $b=1$, $x=m$.
If I'm not mistaking, fixing the iteration level in finding higher solutions but letting $m$ vary instead sould give polynomial parametrisations instead of exponential ones, in case those appeal more to you.
A: As pointed out by Barto, appropriate substitutions can transform the problem into solving a Pell-like, or even a Pell equation. For example, let,
$$a =2bn^2,\quad y = bnu/2$$
and $F_k=\frac{1}{8}ab(a^2+b^2-1) = y^2\,$ transforms into the Pell,
$$u^2-(4n^4+1)b^2 = -1\tag1$$
An initial solution is $u_0 = 2n^2,\; b_0 = 1$ from which we can get an infinite more. For example, the third generation is, 
$$u = 2n^2(3+16n^4),\quad b = 1+16n^4$$
which implies the initial cube,
$$c = n^2(1-4n^2)^2$$
Example: 
Let $n = 2$, so $b=257$, with initial cube $c=30^2=900$, and end cube $900+256=1156$,
$$900^3+901^2+902^3+\dots+1156^3 = 532504^2$$
And so on, for an infinite family of polynomials generated from $u_0,\,b_0$.
