2022/067) Find the no of solutions of equation $\frac{1}{m} + \frac{1}{n} = \frac{1}{143}$ where m , n are distinct positive inetegers

 We have $\frac{1}{m} + \frac{1}{n} = \frac{1}{143}$

or $143n + 143m = mn$

or mn - 143m - 143n = 0

this if of the form $m(n-143) - 143n = 0$

To solve these type of probem add and $143^2$ to both sdes to get

 $m(n-143) - 143(n- 143)= 143^2$

or $(m-143)(n-143) = 143^2= 11^2 * 13^2$

we get the following pairs as solutions (n-143,n-143) = (1,20449), (11, 1859), (13,1573), (121,169),(143,143), (169,121), (1573,13), (1859,11),(20449.1)

or (n,m) = (144, 20592), (11,2002),(13,1716), (264, 312), (286,286), (312,264),(1716,13),(2002,11),(20592.144)

there are 9 pairs out of which 8 are distinct