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 Sum powers do not take five (Posted on 2011-06-30)
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:

n=2

(3+k)(3+k)=(k^2+6k+9)

(3-k)(3-k)=(k^2-6k+9)

2k^2+18=18+16=34

n=3

(3+k)(3+k)(3+k)=(k^3+9k^2+27k+27)

(3-k)(3-k)(3-k)=(-k^3+9k^2-27k+27)

18k^2+54=54+144=198

n=4

(3+k)(3+k)(3+k)(3+k)=k^4+12k^3+54k^2+108k+81

(3-k)(3-k)(3-k)(3-k)=k^4-12k^3+54k^2-108k+81

2k^4+108k^2+162=162+128+864=1154

n=5

(3+k)(3+k)(3+k)(3+k)(3+k)=k^5+15k^4+90k^3+270k^2+405k+243

(3-k)(3-k)(3-k)(3-k)(3-k)=-k^5+15k^4-90k^3+270k^2-405k+243

30k^4+540k^2+486=486+1920+4320=6726

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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