A Very Nice Algebra Problem | Math Olympiad

A Very Nice Algebra Problem | Math Olympiad

x^2 - x - 1 = 0
(x^16 - 1) /(x^8 + 2x^7) = ?

[williamperez-hernandez]

The positive root of x^2 - x - 1= 0, x = gamma =(1+sqrt(5))/2 is related to the Fibonacci Numbers. Namely:
F_n = ((gamma)^n - (-1/gamma)^n)/sqrt(5). For the problem given, factor x^8 in numerator and denominator. The numerator simplifies to the numerator for F_n. The denominator becomes 1+2/gamma = sqrt(5). Therefore the answer is the 8th Fibonacci Number F_8 = 21.
We can generalize this problem, given x^2 - x - 1 = 0, then (x^2n -1)/(x^n+2x^(n-1)) = F_n for n=even. Note if n=odd, we obtain Lucas Number L_n divided by sqrt (5). Note: L_n = (gamma)^n + (-1/gamma)^n.

[PS-mhts]

It's given

x²-x-1=0

i.e., x²=x+1 -- (i)

the denominator of the required expression:

x⁸+2x⁷=x⁷(x+2)

But from (i), we get by adding 1 to both sides:

x²+1=x+2

Therefore, the required denominator=x⁷(x²+1) -- (ii)

Now the numerator = x^16-1, which can be factored as follows:

x^16-1=(x⁸+1)(x⁸-1)
=(x⁸+1)(x⁴+1)(x⁴-1)
=(x⁸+1)(x⁴+1)(x²+1)(x²-1) -- (iii)


From (i), we get: x²-1=x

Thus the
=(x⁸+1)(x⁴+1)(x²+1)x

The denominator (from (ii)) = x⁷(x²+1) -- (iv)


Therefore, the expression to be evaluated is


Cancelling common factors (x and (x²+1)), we get:

(x⁸+1)(x⁴+1)/x⁶

Distributing x⁶ in the denominator as x⁴ and x², we get:


(x⁸+1)(x⁴+1)/x⁶ = [(x⁸+1)/x⁴][(x⁴+1)/x²]

i.e, = (x⁴+1/x⁴)(x²+1/x²) -- (v)


from x²-x-1=0 (which is the given equation), we get:

x²-1=x

dividing throughout by x (which we know can't be zero), we get:

x-1/x=1 - (vi)

squaring both sides of (vi), we get:

x²+1/x²-2=1 => x²+1/x²=1+2=3 -- (vii)


Squaring both sides of (vii), we get:

x⁴+1/x⁴+2=9 => (x⁴+1/x/⁴)=9-2=7 -- (viii)

Using these values in (v), we get our answer=3*7=21

[prabhudasmandal]

The problem is interesting and its solution too. Surface more such problems to help.

[NancyHernandez-joxl]

I would have wanted more of an explanation on how you factored the x^4+1 and the x^4-1

[XBGamerX]

most would go and solve the quadratic given which in fact its the one that gives the golden ratio so it involves irrational solutions and then plug in those solutions into those gigantic exponents on the other expression. but it's wrong, in fact, you should always focus on solving it on some other way like this video, like factoring or adding more equivalent exponents

[nirupamasingh]

V nice step by step explanation. Thank you

[kinshuksinghania]

Properties of higher powers of the golden ratio. Very interesting problem 👍

[andrewgehrett]

Crazy how all that can be simplified down to 9+10

[johnstanley]

Here is a better method use either Cayley-Hamilton or simple long division
and get remainder from dividing by (x^2-x-1)
(x^16-1)/(x^2-x-1) --> 987*x +609
(x^8+2*x^7) --> 47*x+29
hence: (x^16-1)/(x^8+2*x^7) = (987*x+609)/(47*x+29) = 21

[AhmedanterElsayed]

I solved it easily 287 x+ 609 / 47 x + 29
The final answer is 21

[SSR]

Good dame i miss scool. Also i have a question. Math Olimpiad, Math Olimpiad are you seriously. How is this olimphiad we did those kind of exercises in class. in like I don't know 9th ore 10 grade.