Welcome to our community

Be a part of something great, join today!

Value of a

Albert

Well-known member
Jan 25, 2013
1,225
a、b are integers,and

$ax^{17}+ bx^{16}+1$$=Q(x)\times (x^2-x-1)$

where Q(x) is quotient

please find a
 
Last edited:

Albert

Well-known member
Jan 25, 2013
1,225
let :$x^2-x-1=0,\,\,\therefore x^2=x+1, \,\, and,\,\, 987x^2-987x-987=0$

$x^4=3x+2,\,\, x^8=21x+13,\,\,x^{16}=987x+610,\,\,x^{17}=987x^2+610 x$

$ax^{17}+bx^{16}+1=987ax^2+610ax+987bx+610b+1$

$=987(ax^2)+987x(b+\dfrac {610 a}{987})+987(\dfrac {610b+1}{987})$

now we can see :

a=987, 610+b=-987 (and, $ 610b+1=-987^2$)

that is a=987 , b=-1597, $Q(x)=\dfrac{987 x^{17}-1597x^{16}+1}{x^2-x-1}$
it is very tedious to find Q(x)
 
Last edited:

jakncoke

Active member
Jan 11, 2013
68
This can also be done if you recognize that the relationship between the golden ratio and fibonacci sequence

$F(n) = \frac{\phi^n - (1-\phi)^n}{\sqrt{5}}$

solving $x^2-x-1=0$, gives you 2 solutions $\phi, 1-\phi$, (which are related to the fib sequence by the above function F(n).

Using the above relation, we note that the left side shoud equal 0, if we input these roots.a*1597 + 987*b = 0

$a(\phi)^{17} + b(\phi)^{16} + 1 = 0$
$a(1-\phi)^{17} + b(1-\phi)^{16} + 1 = 0$

subtract first from second to get
$a*(\phi^{17} - (1-\phi)^{17}) + b*(\phi^{16} - (1-\phi)^{16}) = 0$
divide by $\sqrt{5}$ to get
$\frac{a*(\phi^{17} - (1-\phi)^{17})}{\sqrt{5}} + \frac{b*(\phi^{16} - (1-\phi)^{16})}{\sqrt{5}} = 0$

We see that a*F(17) + b*F(16) = 0, where F is the fibonacci function
F(17) = 1597
F(16) = 987

so

a*1597 + 987*b = 0

Since gcd(1597,987) = 1
a = 987, b = -1597