# Difference between revisions of "1969 Canadian MO Problems/Problem 1"

## Problem

Show that if $a_1/b_1=a_2/b_2=a_3/b_3$ and $p_1,p_2,p_3$ are not all zero, then $\left(\frac{a_1}{b_1} \right)^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}$ for every positive integer $n.$

## Solution

Instead of proving the two expressions equal, we prove that their difference equals zero.

Subtracting the LHS from the RHS, $0=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}-\frac{a_1^n}{b_1^n}.$

Finding a common denominator, the numerator becomes $b_1^n(p_1a_1^n+p_2a_2^n+p_3a_3^n)-a_1^n(p_1b_1^n+p_2b_2^n+p_3b_3^n)=p_2(a_2^nb_1^n-a_1^nb_2^n)+p_3(a_3^nb_1^n-a_1^nb_3^n)=0.$ (The denominator is irrelevant since it never equals zero)

From $a_1/b_1=a_2/b_2,$ $a_1^nb_2^n=a_2^nb_1^n.$ Similarly, $a_1^nb_3^n=a_3^nb_1^n$ from $a_1/b_1=a_3/b_3.$

Hence, $a_2^nb_1^n-a_1^nb_2^n=a_3^nb_1^n-a_1^nb_3^n=0$ and our proof is complete.

 1969 Canadian MO (Problems) Preceded byFirst question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed byProblem 2

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