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Multiples of 8 plus 1

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Multiples of 8 plus 1

any odd integer can be written as 2k+1, k is an integer >=1 for this problem => (2k+1)^2 = 4k^2 +4k+1 = 4k(k+1) + 1 k, k+1 are consective integers, hence one is even =>k(k+1) = 2n =>4k(k+1) + 1 = 8n +1 and we are done

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