It happens only at n=125 20x125 = 2500 (50x50) 5x125 + 275 = 900 (30x30) So sum of all such positive numbers = 125 ( its the only one).
First 17 positive integers (1..17) are rearranged into a sequence such that the sum of any two adjacent terms is a perfect square. What is the sum of the first and last terms of this sequence?
If 1^3 + 2^3 + 3^3 = m^2 where m is also an integer. What are the next three consecutive positive integers such that the sum of their individual cubes is equal to a perfect square?
