20050916, 17:01  #12  
Nov 2003
2^{2}×5×373 Posts 
Quote:
We need k^6 + c1 k^5 + c2 k^4 + ... = 64 

20050916, 18:58  #13  
May 2003
Warsaw
3×5 Posts 
Quote:
There must be satisfied z^{6} ( (z + k/z)^{6} + c_{1} (z + k/z)^{5} + c_{2} (z + 2/z)^{4} + ... ) = z^{12} + ...  448z  64 The only constant term on the left side is k^{6}. The others are divisible by z. 

20050916, 20:07  #14  
Nov 2003
2^{2}·5·373 Posts 
Quote:


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