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Integer Pairs (Posted on 2013-05-30) Difficulty: 3 of 5
Find all pairs (x,n) of positive integers such that
xn + 2n + 1 is a divisor of xn+1 + 2n+1 + 1

See The Solution Submitted by K Sengupta    
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Some Thoughts computer exploration | Comment 1 of 2

   10   for Tot=2 to 999
   20     for X=1 to Tot-1
   30       N=Tot-X
   40       Lhs=X^N+2^N+1:Rhs=X^(N+1)+2^(N+1)+1
   50       if Rhs @ Lhs=0 then print X;N,Lhs;Rhs
   60     next
   70   next

finds only

 4  1    7  21
 11  1   14  126

for x, n, first expression, second expression.


  Posted by Charlie on 2013-05-30 17:43:21
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