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Tenth Philippine Mathematical Olympiad

Number theory Difficulty 5.7 AIME, harder Prove it Philippines

Problem:
How many ordered pairs (x,y)(x, y) of positive integers, where x<yx < y, satisfy the equation
1x+1y=12007 \frac{1}{x} + \frac{1}{y} = \frac{1}{2007}

Solution

Solution:
We can rewrite the given equation into
(x−2007)(y−2007)=20072=34⋅2232 (x - 2007)(y - 2007) = 2007^{2} = 3^{4} \cdot 223^{2}
Since x<yx < y, we have x−2007<y−2007x - 2007 < y - 2007. It follows that
−2007<x−2007<2007or∣x−2007∣<2007 -2007 < x - 2007 < 2007 \quad \text{or} \quad |x - 2007| < 2007
Thus, we have ∣y−2007∣>2007|y - 2007| > 2007.
x−2007y−2007134⋅2232333⋅22323232⋅2232333⋅223234223222334⋅2233⋅22333⋅223 \begin{array}{rl} x - 2007 & y - 2007 \\ \hline 1 & 3^{4} \cdot 223^{2} \\ 3 & 3^{3} \cdot 223^{2} \\ 3^{2} & 3^{2} \cdot 223^{2} \\ 3^{3} & 3 \cdot 223^{2} \\ 3^{4} & 223^{2} \\ 223 & 3^{4} \cdot 223 \\ 3 \cdot 223 & 3^{3} \cdot 223 \end{array}
For every pair of values of x−2007x - 2007 and y−2007y - 2007 in the above table, there is a corresponding pair of xx and yy. Thus, there are seven such ordered pairs.

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