Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.
Find a if a and b are integers such that \(x^{2}-x-1\) is a factor of \(ax^{17}+bx^{16}+1\).
Integers
Digits
Sets
Answer: is 987.
AIME I, 1988, Question 13
Elementary Number Theory by David Burton
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
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
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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