Not infinite

Find the number of integral ordered pairs ( x , y ) (x,y) such that:

1 x + 1 y = 1 100 \frac {1}{x}+\frac {1}{y}=\frac {1}{100}


The answer is 49.

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1 solution

The given equation can be rewritten as x + y x y = 1 100 x y = 100 ( x + y ) x y 100 x 100 y = 0 ( x 100 ) ( y 100 ) = 10000 \dfrac{x + y}{xy} = \dfrac{1}{100} \Longrightarrow xy = 100(x + y) \Longrightarrow xy - 100x - 100y = 0 \Longrightarrow (x - 100)(y - 100) = 10000 .

Now since 10000 = 2 4 5 4 10000 = 2^{4}5^{4} this value has ( 4 + 1 ) ( 4 + 1 ) = 25 (4 + 1)(4 + 1) = 25 positive integer divisors, and thus 2 × 25 = 50 2 \times 25 = 50 integer divisors in general, (i.e., a negative divisor mirroring each positive divisor). So we can represent 10000 10000 as a pairwise product a b ab in 50 50 distinct ways, assigning a = x 100 , b = y 100 a = x - 100, b = y - 100 , which in turn gives us 50 50 distinct pairs ( x , y ) (x,y) . However, one of these pairs is ( a , b ) = ( 100 , 100 ) ( x , y ) = ( 0 , 0 ) (a,b) = (-100,-100) \Longrightarrow (x,y) = (0,0) , which would not satisfy the original equation. Thus we are left with a final total of 49 \boxed{49} ordered integer pair solutions to the given equation.

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