Another easy one

Number Theory Level pending

How many different pairs of positive integers (x, y) are solutions of the equation x 2 + 3 y 2 = 1997 x^{2} +3y^{2} = 1997 ?


The answer is 0.

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1 solution

_ Just notice, that every square can have remainder 0 or 1 (modulo 3) . 1997 has the remainder 2 (modulo 3) and 3y^{2} = 0 (modulo 3). Now, let's add all of the remainders: {0,1} + {0} = 2, which is impossible for integers_

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