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1969 Canadian MO Problems/Problem 1

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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_2b_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 by
First question
1 2 3 4 5 6 7 8 9 10 Followed by
Problem 2


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