Steps: 1) f injectif 2) f(1)=1 3)ff(m))=m 4)f is multiplif 5)f(p) =p for all p prime with 2024 6)finally !!! Prouve {f(2),f(11),f(23)}={2,11,23} Remarque : f(m) = m is on of the solution but not the only

How to find ALL functions? This is an interesting problem!

Just chose the image of 2, 11, 23 after for all p f(p) = p (p prime) and we have f is multiplicatif so Q. E. D

Of course the image of 2 are one of the following value 2,3,11 and the same for 3 and 11

Let's supply a complete proof of this nice problem. Let $P(m,n)$ denotes the assertion that $f(f(m)f(n)) = mn$. Note that $f$ is injective and $P(1,1)$ gives $f(f(1)^2) = 1$. Now let $m_0:=f(1)^2$. Then $P(m_0,m_0)$ gives $f(1) = m_0^2 = f(1)^4$, yielding $f(1)=1$. This, together with $P(m,1)$ gives $f(f(m))=m$ and $P(f(m),f(n))$ gives $f(mn)=f(m)f(n)$. So, $f$ is multiplicative; it therefore suffices to recover $f$ on primes. We next claim $f(p)$ is a prime for every prime $p$. Indeed, suppose there is a prime $p$ such that $f(p)=qr$ where $q,r>1$. Then $p=f(f(p))=f(qr)=f(q)f(r)$, a contradiction: $f(q),f(r)>1$ since $q,r\ne 1$, $f(1)=1$ and $f$ is injective. Now, we prove $f(p)=p$ for every prime $p$ where $(p,2024)=1$. To that end, we use the second condition. Take any prime $p\notin \{2,11,23\}$. In particular, $(p,2024)=1$ and there exists an $a$ such that $pa\equiv 1\pmod{2024}$. Using Dirichlet's theorem, there exists infinitely many primes congruent to $a$ modulo 2024, let $q_1,q_2$ be two distinct such primes (other than $p$ itself). Then, $f(pq_1)=pq_1$, $f(pq_2)=pq_2$. On the other hand, $f(pq_1)=f(p)f(q_1)$ and $f(pq_2)=f(p)f(q_2)$. So, $p=(pq_1,pq_2)=(f(p)f(q_1),f(p)f(q_2)) = f(p)(f(q_1),f(q_2)) = f(p)$ as $f(q_1)$ and $f(q_2)$ are coprime (recall that $f(r)$ is a prime for any $r$, $f$ is injective and $q_1\ne q_2$). So, $f(p)=p$ for every $p\notin\{2,11,23\}$. Lastly, we note that $f(p)$ is a prime for all $p\in\{2,11,23\}$. As $f(p)=p$ for $p\notin\{2,11,23\}$, we obtain that $(f(2),f(11),f(23))$ is a permutation of $(2,11,23)$. Any such permutation, together with $f(p)=p$ for all $p\notin\{2,11,23\}$, together with the multiplicative property of $f$ uniquely reveals $f$. The end.

grupyorum wrote: Let's supply a complete proof of this nice problem. Let $P(m,n)$ denotes the assertion that $f(f(m)f(n)) = mn$. Note that $f$ is injective and $P(1,1)$ gives $f(f(1)^2) = 1$. Now let $m_0:=f(1)^2$. Then $P(m_0,m_0)$ gives $f(1) = m_0^2 = f(1)^4$, yielding $f(1)=1$. This, together with $P(m,1)$ gives $f(f(m))=m$ and $P(f(m),f(n))$ gives $f(mn)=f(m)f(n)$. So, $f$ is multiplicative; it therefore suffices to recover $f$ on primes. We next claim $f(p)$ is a prime for every prime $p$. Indeed, suppose there is a prime $p$ such that $f(p)=qr$ where $q,r>1$. Then $p=f(f(p))=f(qr)=f(q)f(r)$, a contradiction: $f(q),f(r)>1$ since $q,r\ne 1$, $f(1)=1$ and $f$ is injective. Now, we prove $f(p)=p$ for every prime $p$ where $(p,2024)=1$. To that end, we use the second condition. Take any prime $p\notin \{2,11,23\}$. In particular, $(p,2024)=1$ and there exists an $a$ such that $pa\equiv 1\pmod{2024}$. Using Dirichlet's theorem, there exists infinitely many primes congruent to $a$ modulo 2024, let $q_1,q_2$ be two distinct such primes (other than $p$ itself). Then, $f(pq_1)=pq_1$, $f(pq_2)=pq_2$. On the other hand, $f(pq_1)=f(p)f(q_1)$ and $f(pq_2)=f(p)f(q_2)$. So, $p=(pq_1,pq_2)=(f(p)f(q_1),f(p)f(q_2)) = f(p)(f(q_1),f(q_2)) = f(p)$ as $f(q_1)$ and $f(q_2)$ are coprime (recall that $f(r)$ is a prime for any $r$, $f$ is injective and $q_1\ne q_2$). So, $f(p)=p$ for every $p\notin\{2,11,23\}$. Lastly, we note that $f(p)$ is a prime for all $p\in\{2,11,23\}$. As $f(p)=p$ for $p\notin\{2,11,23\}$, we obtain that $(f(2),f(11),f(23))$ is a permutation of $(2,11,23)$. Any such permutation, together with $f(p)=p$ for all $p\notin\{2,11,23\}$, together with the multiplicative property of $f$ uniquely reveals $f$. The end. NOT any such permutation will work for $(2,11,23)$ You should have one of the numbers fixed to obtain $f(f(m))=m$