Do there exist three integers each greater than one such that the square of each less one is divisible by both the others?
Answer
No.
Solution
Take x > y > z. Then x divides y2 - 1, so x and y are coprime. Both divide z2-1, so their product xy must divide z2 - 1. But xy > z2 > z2-1. Contradiction.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
12 January 2004
Last corrected/updatd 12 Jan 04