$3(2a+m+1)^2=11m^2+1$ has infinite many solutions . And sum is $(a+1)^2+...+(a+m)^2=m^3$

RagvaloD wrote: $3(2a+m+1)^2=11m^2+1$ has infinite many solutions . Why?

$a=21,m=47$ is a particular solution. $22^2+\cdots+68^2=47^3$

jestrada wrote: $2^2+3^2+4^2+5^2+6^2=125$ Are you sure?

I made a mistake expanding $(a+1)^2+\cdots+(a+m)^2$. Fixed. $$(a+1)^2+\cdots+(a+m)^2=ma^2+m(m+1)a+\frac{m(m+1)(2m+1)}6=m^3$$$$\iff a^2+(m+1)a+\frac{(m+1)(2m+1)}6$$$$\iff 6a^2+6am-4m^2+3m+6a+1=0$$After multiplying by 2, this can be rewritten as $3(2a+m+1)^2=11m^2+1$ which is a Pell-type equation; if it has a nontrivial solution it has infinite solutions.

$x_0=2,x_1=2,x_n=46x_{n-1}-x_{n-2}$ $y_0=-1,y_1,=1,y_n=46y_{n-1}-y_{n-2}$ $m=y_n, a=\frac{x_n-y_n-1}{2}$ Some first values :$(a,m)=(21,47),(988,2161)$