Bijective Functions Modulo N

For any integer n , n, a function f : Z Z f:\mathbb{Z} \rightarrow \mathbb{Z} is called bijective modulo n n if the remainders of { f ( 0 ) , f ( 1 ) , , f ( n 1 ) } \{f(0), f(1), \cdots , f(n-1) \} upon division by n n are equivalent to { 0 , 1 , , n 1 } \{0, 1, \cdots , n-1 \} in some order.

A pair of integers ( a , b ) (a,b) is called friends with n n if the function f ( x ) = a x 3 + b x f(x) = ax^3 + bx is bijective modulo n . n. Find the number of primes p < 50 p < 50 which satisfy the following property:

  • If ( a , b ) (a,b) is any pair of integers which is friends with p , p, it is friends with n n for infinitely many integers n . n.

Details and assumptions

  • As an explicit example, consider the function f ( x ) = 7 x f(x) = 7x and take n = 4. n=4. We have { f ( 0 ) , f ( 1 ) , f ( 2 ) , f ( 3 ) } = { 0 , 7 , 14 , 21 } . \{ f(0) , f(1), f(2), f(3) \} = \{0, 7, 14, 21\}. Upon division by 4 , 4, these numbers leave remainders { 0 , 3 , 2 , 1 } \{0, 3, 2, 1\} respectively, which is equivalent to { 0 , 1 , 2 , 3 } \{0, 1, 2, 3\} in another order. Thus, f ( x ) = 7 x f(x) = 7x is bijective modulo n = 4. n=4.

  • You may refer to a list of primes .

  • This problem is not entirely original.


The answer is 6.

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