### Third Millennium International Mathematical Olympiad 2009

Grade 9

Problem 3

Find all pairs of natural numbers A and B for which ^{B} - B^{2})^{2} - (2^{B})^{2} = 2009 |

Solution by Clement Lau, Temple City, CA.

|Up| |Contact| |Front page| |Contents|

Copyright © 1996-2018 Alexander Bogomolny

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

Answer: 2009=7×7×41

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

(A^{B} - B^{2} + 2^{B})(A^{B} - B^{2} - 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

- (A
^{B}- B^{2}+ 2^{B}) = 49 and - (A
^{B}- B^{2}- 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:

- (A
^{B}- B^{2}+ 2^{B}) = 287 and - (A
^{B}- B^{2}- 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:

- (A
^{B}- B^{2}+ 2^{B}) = 2009 and - (A
^{B}- B^{2}- 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.

|Up| |Contact| |Front page| |Contents|

Copyright © 1996-2018 Alexander Bogomolny

69953758