After some calculation, we get that we've to find out an integer $m$ such that \[\sqrt{11m^2-10}\]
is a positive integer greater than $6$ and $m=13$ satisfies the condition
.
So we get \[\sum_{k=38}^{48}k^2=143^2\]
So we're done
!
is a positive integer greater than $6$ and $m=13$ satisfies the condition
.So we get \[\sum_{k=38}^{48}k^2=143^2\]
So we're done
!