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Summing the pairs (Posted on 2019-03-01) Difficulty: 3 of 5
Let (a, b) be a pair of integers such that ax17+bx16+1 is divisible by x2-x-1.

Find the sum of all such pairs.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Comment 1 of 1
ax17+bx16+1=p(x)        x2-x-1=r(x)
p(x)=q(x)*r(x)
q(x)=q1x15+q2x14+q3x13+ (...) +q14x2+q15x1+q16
p(0)=1=q16*(-1)  => q16=-1

For:
x^17 => q1=a
x^16 => q2-q1=b => q2=a+b
x^15 => q3-q2-q1=0 => q3=q2-q1

x^n => q(18-n)=q(18-n-1)+q(18-n-2)

x^2 => q14-q15-1=0
x^1 => -q15+1=0 => q15=1
x^0 => q16=-1 

As we have q16 & q15 we can calculate q14, and then going up q13 and so on.

q16=-1
q15=1 
q14=0
q13=-1
q12=1
q11=-2
q10=3
q9=-5
q8=8
q7=-13
q6=21
q5=-34
q4=55
q3=-89
q2=a+b
q1=a
 
Then:

2*a+b=-89  and
a+b-89=55  

a=-233 b=377 

a+b=144

q(n) follows Fibonacci numbers and also a e b are Fibonacci numbers

Edited on March 2, 2019, 5:48 pm
  Posted by armando on 2019-03-02 17:41:42

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