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Divisible by 7 (Posted on 2007-02-15) Difficulty: 3 of 5
Prove that if A = ½(15+√197), then 7 is a factor of [A^n] for all positive integer values of n, where [w] denotes the greatest integer less than or equal to w.

  Submitted by Dennis    
Rating: 4.0000 (1 votes)
Solution: (Hide)
If B=.5(15-sqrt(197)), then A and B both satisfy x^2-15x+7=0. Let a(1),a(2),a(3), ... be an infinite sequence defined by a(1)=1, a(2)=15, and a(n)=15a(n-1)-7a(n-2). Now a(n)=a(n-1) mod 7 --> a(n)=1 mod 7.

If x^2-15x+7=0 then x^n=a(n)x-7a(n-1) for integer values of n>1. Proof (by induction): n=2 --> x^2=15x-7. Assume x^n=a(n)x-7a(n-1) --> x^(n+1)=a(n)x^2-7a(n-1)x = a(n)(15x-7)-7a(n-1)x = x(15a(n)-7a(n-1))-7a(n) = a(n+1)x-7a(n).

So A^n+B^n=a(n)A-7a(n-1)+a(n)B-7a(n-1) = a(n)(A+B)-14a(n-1) = 15a(n)-14a(n-1) = a(n) mod 7 = 1 mod 7 --> there exists positive integer k such that A^n+B^n = 7k+1 --> A^n=7k+1-B^n. But 0 < B < 1 --> 0 < B^n < 1 --> [A^n]=7k for n>1. Since [A]=14, the proof is complete.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionArek2007-03-05 07:26:24
Hints/Tipsre(3): Look for the conjugateDennis2007-02-26 12:14:00
Some Thoughtsre(2):Look 4 the conjugate.... still openAdy TZIDON2007-02-26 08:16:13
Some Thoughtsre:Look 4 the conjugateDennis2007-02-25 21:39:20
SolutionLOOK 4 THE CONJUGATEAdy TZIDON2007-02-25 16:23:23
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