1969 Canadian MO Problems/Problem 1

Problem

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

Solution

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

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

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

From $\displaystyle a_1/b_1=a_2b_2,$ $\displaystyle a_1^nb_2^n=a_2^nb_1^n.$ Similarly, $\displaystyle a_1^nb_3^n=a_3^nb_1^n$ from $\displaystyle a_1/b_1=a_3/b_3.$

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

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1969 Canadian MO Problems/Problem 1

Problem

Solution