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Problem on Fibonacci sequence | AIME I, 1988 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.

Fibonacci sequence Problem - AIME I, 1988

Find a if a and b are integers such that $x^{2}-x-1$ is a factor of $ax^{17}+bx^{16}+1$.

• is 107
• is 987
• is 840
• cannot be determined from the given information

Key Concepts

Integers

Digits

Sets

AIME I, 1988, Question 13

Elementary Number Theory by David Burton

Try with Hints

First hint

Let F(x)=$ax^{17}+bx^{16}+1$

Let P(x) be polynomial such that

$P(x)(x^{2}-x-1)=F(x)$

constant term of P(x) =(-1)

now $(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+....+c_{15}x-1)$ where $c_{i}$=coefficient

Second Hint

comparing the coefficients of x we get the terms

since F(x) has no x term, then $c_{15}$=1

getting $c_{14}$

$(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+....+c_{15}x-1)$

=terms +$0x^{2}$ +terms

or, $c_{14}=-2$

proceeding in the same way $c_{13}=3$, $c_{12}=-5$, $c_{11}=8$ gives a pattern of Fibonacci sequence

Final Step

or, coefficients of P(x) are Fibonacci sequence with alternating signs

or, a=$c_1=F_{16}$ where $F_{16}$ is 16th Fibonacci number

or, a=987.

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