a) Prove that for each positive integer n there is a unique positive integer an such that (1+√5)n=√an+√an+4n.b) Prove that a2010 is divisible by 5⋅42009 and find the quotient
Problem
Source: 2010 Saudi Arabia IMO TST V p2
Tags: number theory, algebra, Sequence
28.12.2021 20:44
parmenides51 wrote: a) Prove that for each positive integer n there is a unique positive integer an such that (1+√5)n=√an+√an+4n. Equation (1+√5)n=√x+√x+4n implies (1+√5)n−√x=√x+4n And so (squaring) (1+√5)2n−2√x(1+√5)n+x=x+4n And so √x=(1+√5)2n−4n2(1+√5)n) =(√5+1)n−(√5−1)n2 And so an=b2n where bn=(√5+1)n−(√5−1)n2 Looking at bn, we get b0=0, b1=1 and bn+2=2√5bn+1−4bn A simpl;e induction proves that b2k+1 is an integer and that b2k is the product of √5 by an integer. And so an=b2n always is a positive integer. Q.E.D.
28.12.2021 21:28
Interesting If we change 5 by p a prime does exist a unique positive integer y such that (1+√p)n=√y+√y+(p−1)n ?
29.12.2021 08:24
Moubinool wrote: Interesting If we change 5 by p a prime does exist a unique positive integer y such that (1+√p)n=√y+√y+(p−1)n ? Exactly the same computation gives an=b2n where bn=(√p+1)n−(√p−1)n2 Looking again at bn, we get b0=0, b1=1 and bn+2=2√pbn+1−(p−1)bn The same simple induction proves that b2k+1 is an integer and that b2k is the product of √p by an integer. And so b2n always is an integer. Q.E.D.