Problem: a x 5 + b y 5 a x 5 + b y 5 a , b , x a , b , x y y
a x + b y = 3 , a x 2 + b y 2 = 7 , a x 3 + b y 3 = 16 , a x 4 + b y 4 = 42 a x + b y = 3 , a x 2 + b y 2 = 7 , a x 3 + b y 3 = 1 6 , a x 4 + b y 4 = 4 2
Solution:
For n = 1 n = 1 n = 2 n = 2
( a x n + 1 + b y n + 1 ) ( x + y ) − ( a x n + b y n ) x y = a x n + 2 + b y n + 2 (1) ( a x n + 1 + b y n + 1 ) ( x + y ) − ( a x n + b y n ) x y = a x n + 2 + b y n + 2 ( 1 )
yields the equations
7 ( x + y ) − 3 x y = 16 and 16 ( x + y ) − 7 x y = 42 7 ( x + y ) − 3 x y = 1 6 and 1 6 ( x + y ) − 7 x y = 4 2
Solving these last two equations simultaneously, one finds that
x + y = − 14 and x y = − 38 (2) x + y = − 1 4 and x y = − 3 8 ( 2 )
Applying ( 1 ) ( 1 ) n = 3 n = 3
a x 5 + b y 5 = ( 42 ) ( − 14 ) − ( 16 ) ( − 38 ) = − 588 + 608 = 20 a x 5 + b y 5 = ( 4 2 ) ( − 1 4 ) − ( 1 6 ) ( − 3 8 ) = − 5 8 8 + 6 0 8 = 2 0 ​
Note: From ( 2 ) ( 2 ) x x y y x = − 7 ± 87 x = − 7 ± 8 7 ​ y = − 7 ∓ 87 y = − 7 ∓ 8 7 ​ a = 49 76 ± 457 6612 87 a = 7 6 4 9 ​ ± 6 6 1 2 4 5 7 ​ 8 7 ​ b = 49 76 ∓ 457 6612 87 b = 7 6 4 9 ​ ∓ 6 6 1 2 4 5 7 ​ 8 7 ​
The problems on this page are the property of the MAA's American Mathematics Competitions