194040 has factors of 2,2,2,5 and 4951 all of which are prime numbers and their sum is 4970
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?
If a, b, c and d are distinct pairwise co-prime positive integers such that a^2 + b^2 = c^2 + d^2, find the lowest possible value of a + b + c + d?
