Question

Find the number of ordered pairs (x;y) of positive integers satisfying 1/x + 1/y = 1/(2021 ^ 17)

If x ≠ y then pairs (x;y) and (y;x) are considered to be different.

Answer 1

$\frac{x+y}{xy}=\frac{1}{2021^{17}}$

$xy-2021^{17}(x+y)=0$

Adding $(2021^{17})^2$ both sides

$xy-2022^{17}(x+y)+(2021^{17})^2=(2021^{17})^2$

$(x-2021^{17})(y-2021^{17})=(2021^{17})^2$

Let $x-2021^{17}=A$ and $y-2021^{17}=B$

$\therefore\>AB=(2021^{17})^2$

$(2021^{17})^2=43^{34}×47^{34}$

Number of factors $=$ $(34+1)(34+1)$

$=1225$

In one case $A=B$

$\therefore$ Number of ordered pairs $=1224$