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    The Fibonacci sequence  {f(n)}=1,1,2,3,5,8,13,...   is defined by the
conditions   f(1)=f(2)=1,   f(n+2)=f(n)+f(n+1),  for  n>=1.
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Problem 1.  Prove that   
                      W(n) = f(n+2)f(n)-(f(n+1))^2=(-1)^(n+1),       n>=1.
Proof.  By induction in  n.  For  n=1,  the equality holds:  
    W(1) =  2*1-1^2=(-1)^2  (both sides =1).
Step induction.  Assume that it is true for  n<=N.  Then

W(N+1) = f(N+3)f(N+1)-(f(N+2))^2 = 
   =  [f(N+2)+f(N+1)]f(N+1) - (f(N+2))^2  =
   =  f(N+2)[f(N+1)-f(N+2)]  + (f(N+1))^2 = 
   =  f(N+2)(-f(N)) + (f(N+1))^2 = -W(N) = -(-1)^(N+1) = (-1)^(N+2),
   completing the proof.
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Problem 2.  Prove that   (f(n), f(n+1)) = 1,   for   n„1.
Proof.  By induction in  n.  True for  n=1:  (1,1)=1.
Step induction.  Assume that it is true for  n¾N.  Then
   (f(N+1), f(N+2))  =  (f(N+1), f(N+2) - f(N+1)) =
  =  (f(N+1), f(N)) =1,  completing the proof.
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Problem 3.  (f(n+k), f(n)) | f(k),  n,k>=1.  
  We have
       f(n+2)=f(n)+f(n+1);
       f(n+3)=f(n)+2f(n+1);
       f(n+4)=2f(n)+3f(n+1);
       f(n+5)=3f(n)+5f(n+1);
and in general
       f(n+k)=f(k-1)f(n)+f(k)f(n+1).
(The last identity is easily  proved,  for any fixed n,
   by induction in  k.)
It follows:
      (f(n+k), f(n)) =  (f(k-1)f(n)+f(k)f(n+1), f(n)) =
       =  (f(k)f(n+1), f(n)) =  (f(k), f(n))
which is a divisor of  f(k).  END OF PROOF.
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