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12 &14

x^2+(x+2)^2=340 ===> positive solution is x=12 numbers are 12 &14

x^2+(x+2)^2=340 x^2+x^2+4x+4=340 2x^2+4x-336=0 x^2+2x-168=0 x^2+14x-12x-168 x(x+14)-129x+14=0 (x+14)(x-12)=0 x=12 or-14 x=12 as x is positive So the numbers are 12 and 14

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?

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?