Problem:
Let f1​(x)=32​−3x+13​, and for n≥2, define fn​(x)=f1​(fn−1​(x)). The value of x that satisfies f1001​(x)=x−3 can be expressed in the form nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Note that for xî€ =32​ or −31​,
​f1​(x)=32​−3x+13​=9x+36x−7​f2​(x)=f1​(f1​(x))=9⋅(9x+36x−7​)+36⋅(9x+36x−7​)−7​=9x−6−3x−7​, and f3​(x)=f1​(f2​(x))=9⋅(9x−6−3x−7​)+36⋅(9x−6−3x−7​)−7​=x.​
Thus it follows that f3k​(x)=x for all positive integers k. Thus f1001​(x)= f2​(x)=9x−6−3x−7​=x−3, which is equivalent to 9x2−30x+25=0, and the only solution to this quadratic is x=35​. Thus m+n=5+3=8​.
Challenge: Show that if p+q=1 and f1​(x)=p−x+q1​, then f3​(x)=x. The given problem highlights a special case of this fact in which p=32​ and q=31​.
The problems on this page are the property of the MAA's American Mathematics Competitions