$X^2 + 2011 = Y^2$ where X and Y are both positive integers. Find value of 3X + Y

The answer is 4021.

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$(Y - X)( Y + X) = 2011$

Since 2011 is prime, we have: $Y - X = 1 \quad and \quad Y + X = 2011$ Then $Y = 1006 \quad and \quad X = 1005$