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
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Posted by broll
on 2011-07-01 01:57:49 |
Possible solution
