Cycling Through The Polynomial Shuffle

Suppose f ( x ) f(x) is a polynomial with integer coefficients which is bijective modulo 49 49 , which means that x ≢ y ( m o d 49 ) f ( x ) ≢ f ( y ) ( m o d 49 ) . x\not \equiv y \pmod {49} \Leftrightarrow f(x)\not \equiv f(y) \pmod {49}. We define the composition powers f ( i ) ( x ) f^{(i)}(x) by setting f ( 1 ) = f f^{(1)}=f and f ( i + 1 ) ( x ) = f ( f ( i ) ( x ) ) . f^{(i+1)}(x)=f(f^{(i)}(x)). For a given f , f, its order (in modulo 49) is defined as the smallest positive integer n , n, such that f ( n ) ( x ) x ( m o d 49 ) f^{(n)}(x)\equiv x \pmod {49} for all integers x . x.

What is the largest possible order of f ? f?


The answer is 420.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

0 solutions

No explanations have been posted yet. Check back later!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...