Problem 1 of Canadian Mathematical Olympiad 1969

Show that if a1/b1 = a2/b2 = a3/b3 and p1, p2, p3 are not all zero, then

(a1/b1)^n = (p1a1^n + p2a2^n + p3a3^n )/ (p1b1^n + p2b2^n + p3b3^n )

for every positive integer n.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Add a1/b1 ratio to the left onto the already existing equal ratios a1/b1 = a2/b2 = a3/b3, we have

a1/b1 = a1/b1 = a2/b2 = a3/b3

Now raise all to the n power

(a1/b1)^n = (a1/b1)^n = (a2/b2)^n = (a3/b3)^n

Now multiply both sides of different ratios with equal numbers pís

(a1/b1)^n = p1a1^n /(p1b1^n) = p2a2^n /(p2b2^n ) = p3a3^n /(p3b3^n ) =

(p1a1^n + p2a2^n + p3a3^n )/ (p1b1^n + p2b2^n + p3b3^n )