1st CIS 1992 Problem 7 Find all real x, y such that (1 + x)(1 + x 2 )(1 + x 4 ) = 1+ y 7 , (1 + y)(1 + y 2 )(1 + y 4 ) = 1+ x 7 ?
(x,y) = (0,0) or (-1,-1) Solution If x = y, then clearly x ≠ 1, so we have (1-x 8 ) = (1-x)(1+x 7 ) = 1-x 8 -x+x 7 , so x = 0 or x 6 = 1, whose only real root (apart from the x = 1 we have discarded) is x = -1. That gives the two solutions above. So assume x ≠ y. wlog x > y. So (1+x) > (1+y) and (1+x 7 ) > (1+y 7 ). So we must have (1+x 2 )(1+x 4 )  0, then (1 + x)(1 + x 2 )(1 + x 4 ) > 1 > 1+y 7 , so x  0 and each term on rhs < 0. Contradiction. So there are no more solutions. 1st CIS 1992 © John Scholes jscholes@kalva.demon.co.uk 5 December 2003 Last corrected/updated 5 Dec 03