(In reply to
computer solution by Charlie)
n=47, x=-68 is a solution, as negative integers also have perfect squares.
Let sum n=1 to n (x+n)^2 = n^3
Then 1/6n(2n^2+n(6x+3)+6x^2+6x+1) = n^3, so
1/6(2n^2+n(6x+3)+6x^2+6x+1) = n^2, or
1/12 (n^2 - 1) + 1/4 (n + 2 x + 1)^2 = n^2
{{n == 1, x == -2}, {n == 1, x == 0}, {n == 47, x == -69}, {n == 47, x == 21}, {n == 2161, x == -3150}, {n == 2161, x == 988}} etc.
A189173 gives the value of n but does not specify the consecutive squares.
Edited on August 9, 2019, 11:08 pm
|
|
Posted by broll
on 2019-08-09 22:59:25 |
re: computer solution
