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| − | I just wonder if it's possible to solve this problem with Chinese Remainder Theorem
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| − | First: assuming that <math>GCD(a,b)=1</math>.
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| − | Then quotient is always square <math>mod a</math> and <math>mod b</math> and is less or equal than <math>ab</math> and is not divisible by neither <math>a</math> nor <math>b</math> which implies it's square of integer.
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| − | In case of <math>GCD(a,b) = d>1</math> we can transform quotient to <math>d^2((a_1)^2 + (b_1)^2)/(d^2a_1b_1 + 1)</math> where <math>a_1 = a/d</math> and <math>b_1 = b/d</math> and follow the same reasoning as above.
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| − | It's just an idea without final and rigorous proof yet.
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| − | Am I mistaken?
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| − | Help :)
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