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