Find the Fibonacci Kernel
Python 2, 444 391 372 bytes

Crossed out 444 is still regular 444 ;(

Huge thanks to @Dennis for a whopping -52 -71 bytes!



k=lambda c,a,b:abs(a+c*a-c*b)-c&lt;abs(b-a)&gt;0and k(c,b-c*a,a+b-c*b)or(a,b)
def f(*t):
 a,b=sorted(t);m=b-a+1,0;g=lambda _:min([k(1,*k(0,*s)),m][_!=b:],key=lambda(x,y):abs(x-y))
 if b&lt;0:x,y=f(-a,-b);return-x,-y
 for c in range(-b,b+1):
    for s in(c,a),(a,c):
     x,y=s
     if min(s)&gt;0:
        while y&lt;b:x,y=y,x+y
        m=g(y)
     x,y=s
     while(x!=b)&amp;((x&gt;b)^(b&gt;0)):x,y=y-x,x
     m=g(x)
 return m


Try it online!

The solution can be run by calling f(a, b) on the two input integers. It is based on the idea that if both a and b are within at least one of the same sequence (where a and b are ordered beforehand such that a ≤ b), it follows that there is at least one integer c equivalent to an adjacent value of a in a shared sequence of a and b for which the sequence generated by a and c contains b in it.

Furthermore, if at least one of the two integers is positive, all values of c must be bounded by -b ≤ c ≤ b for it to even be possible to generate the value of b on either side of the starting pair. Thus, the solution simply brute-forces values of c between -b and b that in combination with a are capable of generating b within the sequence, and finds the one for which the difference of the kernel values for a and c are minimal (this is possible because finding the kernel for two adjacent numbers in a sequence is trivial).

If neither a nor b is positive, the solution simply negates both and returns the negative of the kernel generated for the negated pair.