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a(36) = 0, because if a(n) exists then k exists such that k^2 + 36 = m^2 where k^2, 36 and m^2 have the same prime signature. Rewriting 36 = m^2 - k^2 = (m - k)*(m + k) and then inspection over divisors of 36 gives no terms. Alternatively checking Pythagorean triplets triples gives the same result. - David A. Corneth, Mar 08 2019
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a(36) = 0, because if a(n) exist exists then k exist exists such that k^2 + 36 = m^2 where k^2, 36 and m^2 have the same prime signature. Rewriting 36 = m^2 - k^2 = (m - k)*(m + k) and then inspection over divisors of 36 gives no terms. Alternatively checking Pythagorean triplets gives the same result. - David A. Corneth, Mar 08 2019
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If n is even and a(n) > 0 and the exponent of 2 in the factorization of n is the largest in the prime signature then a(n) must be isn't necessarily odd. _Ray Chandler_ found n = 392 as an example where a(n) = 108 is even. (End)
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