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Del

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Multiples of 8 plus 1

I kind of ended up with something similar, but I didn't quite use a formal proof. I noticed right away that 8 was multiplied by a triangular increasing number, or that it was the sum + x, or n(n+1)/2. You can multiply this by 8 and get 4n(n+1) + 1 = x^2. With a little work you can then notice the pattern in x, is that it is 2n+1, or the progression of odd numbers. In summary, 4n(n+1) + 1 = (2n+1)^2

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