Sum of square patterns Can anyone give the name of this pattern
$$136^2+137^2+138^2+139^2+140^2+141^2+142^2+143^2+144^2 =\\ 
145^2+146^2+147^2+148^2+149^2+150^2+151^2+152^2$$

It belongs to a family that starts with the familiar,
$$\begin{align}
&3^2+4^2 = 5^2\\
&10^2+11^2+12^2=13^2+14^2\\
&21^2+22^2+23^2+24^2 = 25^2+26^2+27^2\\
\vdots\\
&a^2 + (a+1)^2 + \dots + a_n^2 = b^2 + (b+1)^2 + \dots + b_{n-1}^2
\end{align}$$
where $b = a_n+1$ and $a =  2n^2-3n+1 = 0, 3, 10, 21, 36, 55, 78, 105, \color{brown}{136},\dots$ Yours was the case of $n=9$ squares. In general, since,
$$a^2+(a+1)^2+\dots+(a+p-1)^2 = (p/6)(6a^2-6a + 6ap + 1 - 3p + 2p^2)$$
then you need to solve an equation of the form,
$$p(6a^2-6a + 6ap + 1 - 3p + 2p^2) = q(6b^2 - 6b + 6b q + 1 - 3q + 2q^2)\tag1$$
in positive integers. A parametric solution to $(1)$ is,
$$a = 2p^2-3p+1$$
$$b = a+p$$
$$q = p-1$$
for any $p$. The "continuous" case $b=a+p$ like the family above also has sporadic solutions like the $a,p,q=4,35,10$ given by bobbym, though these do not seem to belong to a second family.
