Drawing Points and Friendliness

A point ( x 1 , y 1 ) (x_1,y_1) is drawn in the positive coordinate plane with x 1 > y 1 x_1 > y_1 . Then, an infinite sequence of points are drawn, with the following rule: { x n + 1 = x n + y n y n + 1 = x n y n \left\{\begin{array}{l}x_{n+1}=x_n+y_n\\ y_{n+1}=x_n-y_n\end{array}\right.

Suppose a starting point ( x 1 , y 1 ) (x_1,y_1) is called friendly to a coordinate ( a , b ) (a,b) if a point is drawn on ( a , b ) (a,b) sometime in the sequence; that is, ( a , b ) = ( x k , y k ) (a,b)=(x_k,y_k) for some positive integer k k .

Let F ( a , b ) F(a,b) be a function that counts the number of starting points ( x 1 , y 1 ) (x_1,y_1) that are friendly to ( a , b ) (a,b) .

The number of ordered pairs ( p , q ) (p,q) satisfying F ( p , q ) = 2014 F(p,q)=2014 and p + q 2 2 10 p+q \le 2^{2^{10}} can be expressed as a b a^b where a , b a,b are positive integers and a a is minimized. Find the value of a + b a+b .

Details and Assumptions

As much as this looks like a Computer Science problem, it is 100% doable with only a pencil and a paper.


The answer is 34.

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1 solution

Somdutt Goyal
Sep 22, 2014

first of all, the problem should be stated with an extra condition i.e. (x1,y1) is called friendly to a coordinate (a,b) if a point is drawn on (a,b) sometime in the sequence and x1 and y1 should be integres. otherwise F(p,q) will always be infinite and that does not make any sense.

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