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Sum powers do not take five (Posted on 2011-06-30) Difficulty: 3 of 5
α and β are two roots of the equation x2 6x + 1 = 0

Prove that αn+ βn is an integer for every positive integer value of n, and it is not divisible by 5 for any positive integer value of n

No Solution Yet Submitted by K Sengupta    
Rating: 2.5000 (2 votes)

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Possible solution | Comment 2 of 3 |

First Part:                                                                                                                                                                 

x2 - 6x + 1 = 0 has two roots, say:  a=3-2*2^(1/2), b=3+2^(1/2)                                           

I   Let a^n+b^n=F, then                                                  

II  F=(3-2*2^(1/2))^n+(3+2*2^(1/2))^n, 0<n<10 {6,34,198...} A003499 in Sloane        

III We need to show that the F is integral for all n. Take small values, substituting k for the square root:   




















IV It is immediately obvious that the non-integral parts invariably cancel, in that for every non-integral term of (3+k)^n there is an equal but opposite term of (3-k)^n; hence F is always integral.    


Second Part            


Take half the above solutions for 2,3,4,5 etc: {1,3,17,99,577} :this gives x^2 - 8*y^2 = 1 where A001541 in Sloane=x, A001109 in Sloane =y {0,1,6,35,204}.  Multiply out to get our own sequence:                                                                             

I   (x/2)^2-8(y/2)^2=1, giving x^2-8y^2 = 4                              

II  Assume that 5 divides x; then 25z^2=8y^2+4:                           

y = -(25x^2-4)^(1/2)/(2*2^(1/2))                                                      

y =  (25x^2-4)^(1/2)/(2*2^(1/2))                                                    

 and there is no way of getting rid of the square root in the numerator on RHS.       

III  Hence x is never divisible by 5.                            


Edited on July 1, 2011, 2:08 am
  Posted by broll on 2011-07-01 01:57:49

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