Discovering Associative Functions: A Prime Puzzle for Homework

eddybob123
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Homework Statement


Let ##d(n)## denote the least prime factor of a positive integer ##n##, and let ##p## and ##q## be prime numbers. Find all functions ##f## such that ##d(f(p,q))## is associative for all ##p## and ##q##.



Homework Equations


##f:\Bbb{P}\times \Bbb{P}\to \Bbb{P}## is a binary mapping of prime numbers.



The Attempt at a Solution


For clarity, we shall call the function composition ##(d\cdot f)(p,q)## simply ##g(p,q)##
To be honest, I'm not even sure such a function exists, let alone try and find it. My first instinct was to expand it out and try to "force" the solution:
$$g(p,g(q,r))))=g(g(p,q),r))$$
which gives us two cases: either ##g## is surjective or ##p=g(p,q)## and ##g(q,r)=r##.
What do you guys think?
 
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As a quick example of such a function f, let f(p,q) = pq. Then d(f(p,q)) = min(p,q). And g(p,g(q,r)) = g(g(p,q),r) = min(p,q,r).

It's unlikely they intend for the image of f to be the primes (which your post seems to imply) as that would make composing it with d fairly boring...
 
eddybob123 said:

Homework Statement


Let ##d(n)## denote the least prime factor of a positive integer ##n##, and let ##p## and ##q## be prime numbers. Find all functions ##f## such that ##d(f(p,q))## is associative for all ##p## and ##q##.



Homework Equations


##f:\Bbb{P}\times \Bbb{P}\to \Bbb{P}## is a binary mapping of prime numbers.



The Attempt at a Solution


For clarity, we shall call the function composition ##(d\cdot f)(p,q)## simply ##g(p,q)##
To be honest, I'm not even sure such a function exists, let alone try and find it. My first instinct was to expand it out and try to "force" the solution:
$$g(p,g(q,r))))=g(g(p,q),r))$$
which gives us two cases: either ##g## is surjective or ##p=g(p,q)## and ##g(q,r)=r##.
What do you guys think?

What's wrong with f(p,q)=2. Or f(p,q)=min(p,q)?
 
I intend to find an algebraic function of p and q.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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