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How do you solve this equation

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If you mean (a^(2) + b^(2))(c^(2) + d^(2)) = 1993, there are none because if a, b, c, and d are positive integers, so are (a^(2) + b^(2)) and (c^(2) + d^(2)). (a^(2) + b^(2)) and (c^(2) + d^(2)) must be factors of 1993, but 1993 is prime, and since a, b, c, and d are all positive integers, no two positive integers add up to 1.
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~~If you know something about the Gaussian integers, it's clear that there will be a unique solution x, y, 1, 0, up to a permutation of the variables and changes of sign. You just need to find one solution of x\^2 + y\^2 = 1993.~~

Never mind. I didn't notice the variables needed to be positive integers.

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