Let N be the set of positive integers. Find all functions f:N->N that satisfy the equation
fabc-a(abc) + fabc-b(abc) + fabc-c(abc) = a + b + c
for all a, b, c ≥ 2.
(Here f1(n) = f(n) and fk(n) = f(fk-1(n)) for every integer k greater than 1)
(Also note: abc is the product a·b·c and not the concatenation 100a+10b+c)
Suppose we already knew the values of f^m(r), where m, r ranges over all values up to abc -1 and we wanted to extend the list to include the case where m, r range over values up to abc, where a < b < c are distinct primes. There are many ways to do this. Here is an example:
Let the sequence of primes be indexed in increasing order, and suppose that p_{k-1} is the largest prime that occurs among any of the quantities f^m(r) with m, r ranging over all values up to abc -1. We then write:
f^q(abc) = p_{k+q} for q = 1,2..,abc - a
except at the 3 special q values
f^{abc - a}(abc) = f(p_{k + abc - a -1}) = a
f^{abc - a}(abc) = f(p_{k + abc - b -1}) = b
f^{abc - a}(abc) = f(p_{k + abc - c -1}) = c
The idea here is we use that fact that the equation used to define f never makes use of any primes in the domain. Hence we are free to define how f behaves on the primes without risking ever interfering with the defining condition.
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Posted by FrankM
on 2021-08-17 16:23:08 |
Approach for expanding into simple cases
