There are only four squares that are equal to the sum of three consecutive cubes.
Zero is one of them.
Find the other three.
(In reply to
re: solutions by Charlie)
That's what I meant
(x1)^3 + x^3 + (x+1)^3 = y^2
3*x^3 + 6*x = y^2
I found as you did four solutions
How can we prove analytically that no other solutions exist?

Posted by Benny
on 20140130 19:01:02 