Third Millennium International Mathematical Olympiad 2009
Grade 9
Problem 3

Find all pairs of natural numbers A and B for which (AB - B2)2 - (2B)2 = 2009.

Solution by Clement Lau, Temple City, CA.

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Copyright © 1996-2018 Alexander Bogomolny

Find all pairs of natural numbers A and B for which (A^B - B²)² - (2^B)² = 2009.

Answer: 2009=7×7×41

(A^B - B²)² - (2^B)² = 2009 -- difference of squares

(A^B - B² + 2^B)(A^B - B² - 2^B) = 2009

Each of the groupings must be an integer because both A and B are natural numbers. Therefore, the two groupings are made up of the factors of 2009: 7, 7, and 41. Here are the possible values for the groupings: 49, 41; 7, 287; 1, 2009

First possibility

1. (A^B - B² + 2^B) = 49 and
2. (A^B - B² - 2^B) = 41.

So, 2×2^B = 8 -- combination of system of equations after b) is multiplied by -1 = -1

2×2^B = 8
2^B = 4
B = 2

Plugging the value of 2 in as B in equation a) yields 7 as the value for A. So, A = 7 and B = 2.

Second possibility:

1. (A^B - B² + 2^B) = 287 and
2. (A^B - B² - 2^B) = 7.

Following the steps above, 2×2^B = 280. 2^B = 140. Since 140 is not an even power of 2, B cannot be a natural number. This possibility does not yield a valid answer.

Third possibility:

1. (A^B - B² + 2^B) = 2009 and
2. (A^B - B² - 2^B) = 1.

Following the steps above, 2×2^B = 2008. 2^B = 1004. Since 1004 is not an even power of 2, B cannot be a natural number. This possibility does not yield a valid answer.

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Copyright © 1996-2018 Alexander Bogomolny