My solution: Let $ \mathbb{S'} = \{1,2,3,...\} $. We will find all such functions from $\mathbb{S'} $ to itself, and then remove those which do not satisfy our needs in $\mathbb{S} $. Indeed, putting $ a = b = 1 $ gives $ f(1)^2 = f(1) $. So $ f(1) = 1$. Now putting $b=1$ gives $ f(a) = f(a^2) $. Hence it is equivalent to $f(a)f(b) = f(a^2 b^2) = f(ab) $, whose solutions are well known. Out of those, only $f(n) = 1 $(constant function) satisfies our condition, but $1$ does not belong to $\mathbb{S} $. Hence the answer is NO.

I am not too confident with my answer, pls check it!

AdithyaBhaskar wrote: I am not too confident with my answer, pls check it! According to me; you cant use this method. The set of functions you studied has more constraints than the original. So you cant conclude that the required set is a subset of. Example : If I ask you to find all functions $f(x)$ from $\mathbb R^+\to\mathbb R^+$ such that $f(xy)=f(x)+f(y)$ $\forall x,y\in\mathbb R^+$, you'll find for example $\ln x$ But if you extand to $\mathbb R^+\cup\{0\}$, you'll only find $f(x)=0$ $\forall x$ and you cant conclue that no solution exists over $\mathbb R^+$

hajimbrak wrote: Let $S = \{2, 3, 4, \ldots\}$ denote the set of integers that are greater than or equal to $2$. Does there exist a function $f : S \to S$ such that $f (a)f (b) = f (a^2 b^2 )$ for all $a, b \in S$ with $a \ne b$? Let $a,b>2$ such that $a\ne 4b^2$ and $b\ne 4a^2$ $f(2)f(a)f(b)=f(4a^2)f(b)=f(16a^4b^2)=f(2ab)f(2a)$ $f(2)f(a)f(b)=f(a)(4b^2)=f(16a^2b^4)=f(2ab)f(2b)$ So $f(2a)=f(2b)$ $\forall a,b>2$ such that $a\ne 4b^2$ and $b\ne 4a^2$ Let then $a,b>2$ Let $c>\max(4a^2,4b^2)$ : $f(2a)=f(2c)=f(2b)$ and so $f(x)=k$ constant over even numbers $\ge 4$ Then $f(2a)f(2b)=f(16a^2b^2)$ where $a,b>1$ becomes $k^2=k$, impossible over $S$ So no such function.

Prove $f(2)f(k)f(2k^2)=f(2)f(k^2)f(2k^2)$, so $f(k)=f(k^2)$ and we easily get a contradiction.

I think it is easy for APMO not medium. Here my solution. Let $ P (a;b)$ be the assertion of the given functions. By putting $ P (a^2 c^2; a^2)$ and $ P(a^2;b) $ we get that $ f (a^4 c^2)=f (a) f (c) f (a^2)=f (a^2) f (c)$. Clearly there is no such function.

MathPanda1 wrote: Prove $f(2)f(k)f(2k^2)=f(2)f(k^2)f(2k^2)$ Could you kindly show how ?

\begin{align*} f(2^1)f(2^2)f(2^3)f(2^4)f(2^5) &= f(2^3)f(2^4)f(2^5)f(2^6) \\ &= f(2^5)f(2^6)f(2^{14}) \\ &= f(2^{14})f(2^{22}) \\ &= f(2^{72}) \end{align*} and \begin{align*} f(2^1)f(2^2)f(2^3)f(2^4)f(2^5) &= f(2^1)f(2^2)f(2^4)f(2^{16}) \\ &= f(2^2)f(2^4)f(2^{34}) \\ &= f(2^4)f(2^{72}) \end{align*} together imply that $f(2^4) = 1$, so no such function exists.

pco wrote: MathPanda1 wrote: Prove $f(2)f(k)f(2k^2)=f(2)f(k^2)f(2k^2)$ Could you kindly show how ? Sure! Here is how I got $f(2)f(k)f(2k^2)=f(2)f(k^2)f(2k^2)$: $f(2)f(k)f(2k^2) =f(2)f(4k^6) =f(2^2 4^2 k^{12}) =f((2k^2)^2 (4k^4)^2) =f(2k^2)f(4k^4)$ $=f(2k^2)f(2^2 (k^2)^2) =f(2k^2)f(2)f(k^2)$.

Here is a slightly more complicated way, but fairly straightforward as well: First observe that the condition $f(a)f(b)=f(a^2b^2)$ can be used to deduce that $\frac{f(a)}{f(b)}=\frac{f(ka)}{f(kb)}$ for all positive integers $k$ (to confirm this just cross multiply and apply the given condition). Lemma: If $a \mid b$, then $f(a) \mid f(b)$. Proof: We have $\frac{f(ka)}{f(a)}=\frac{f(k^2a)}{f(ka)}$ so if $\frac{f(ka)}{f(a)}=\frac{m}{n}$ where $\gcd(m,n)=1$, then we have that $f(k^ja)=\frac{m^j \cdot f(a)}{n^j}$ so $n^j \mid f(a)$ for all positive integers $j$. However, this is impossible unless $f(a)=0$, an impossibility, or if $n=1$. Thus $n=1$ and so $f(ka)=mf(a)$ where $m$ is an integer and we are done. $\Box$ Now consider two positive integers $a,b$ and suppose that $\gcd(f(a),f(b))=d$. Then we have that $\frac{f(a)}{f(b)}=\frac{f(a^2)}{f(ab)}$ and from our lemma, we know that $f(a),f(b) \mid f(ab)$ so $f(ab)$ must be a multiple of $\frac{f(a)f(b)}{d}$, so let us say that $f(ab)=\frac{kf(a)f(b)}{d}$, and similarly, we have $f(a) \mid f(a^2)$ so $f(a^2)=r f(a)$. Then plugging this into the RHS of our equation we have $\frac{f(a)}{f(b)}=\frac{rd}{kf(b)}$ which yields $r=\frac{kf(a)}{d}$. Thus $f(a^2)=\frac{kf(a)^2}{d}$. Now from our lemma, we have $f(a^2),f(b^2) \mid f(a^2b^2)=f(a)f(b)$. The first instance yields that $\frac{kf(a)}{d} \mid f(b)$; however, since $\gcd(\frac{f(a)}{d},f(b))=1$, this cannot be true unless $\gcd(f(a),f(b))=f(a) \implies f(a) \mid f(b)$. However, since $f(b^2) \mid f(a^2b^2)=f(a)f(b)$, we also get $f(b) \mid f(a)$ so $f(a)=f(b)$ for all $a,b$. This means that $f$ is a constant function of the form $f(n)=C$ for some constant $C$. Then by the condition, we must have $C^2=C \implies C=0,1$ both of which are outside of the possible range of $f$, implying that there are no such functions satisfying the conditions in the problem. $\Box$

My solution: Let $g(x) = f(2^x)$ for brevity, then $g(a)g(b) = g(2(a+b))$. Now: $$g(2)g(3)g(1) = g(2)g(8) = g(20) = g(3)g(7)$$ Hence $g(1)g(2) = g(7)$. Now: $$g(1)g(7) = g(16) = g(2)g(6) = g(1)g(2)^2$$ Hence $g(2)^2 = g(7)$ and $g(1) = g(2)$. Setting $n \geq 3$ we have: $$g(1)g(n) = g(2)g(n) = g(2(n+2)) = g(1)g(n+1)$$. So $g(n)$ is constant for $n \geq 3$. This implies $g(14)^2 = g(3)g(4) = g(14)$, impossible. Hence no such function exists.

For @sicilianfan there is no need for too long solution.

Is this too easy for APMO level?

AdithyaBhaskar wrote: My solution: Let $ \mathbb{S'} = \{1,2,3,...\} $. We will find all such functions from $\mathbb{S'} $ to itself, and then remove those which do not satisfy our needs in $\mathbb{S} $. Indeed, putting $ a = b = 1 $ gives $ f(1)^2 = f(1) $. So $ f(1) = 1$. Now putting $b=1$ gives $ f(a) = f(a^2) $. Hence it is equivalent to $f(a)f(b) = f(a^2 b^2) = f(ab) $, whose solutions are well known. Out of those, only $f(n) = 1 $(constant function) satisfies our condition, but $1$ does not belong to $\mathbb{S} $. Hence the answer is NO. How do you know that f(1) exist when the function is not even defined on 1

My solution: $f(a^4b^6)=f(a) f(ab^3)=f (a^2b^2)f(b)=f(a)f(b)^2 \longrightarrow f(ab^3)=f (b^2)$* now take $b \longrightarrow 2, a\longrightarrow 2a^2$ in*then we get $f(2)^2=f (16a^2)=f (4)f(a) \longrightarrow f(a)$ is constant DONE

Hi everyone , where could I find out from the results of APMO?

Hi I think this solution works too Assuming there exists a function $f: S \rightarrow S$: Given three distinct elements $a,b,c \in S$ we have $\frac{f(a)}{f(b)} = \frac{f(a)*f(c)}{f(b)*f(c)} = \frac{f(a^2*c^2)}{f(b^2*c^2)}$ And also $f(a^4b^4) = f(a^2b)*f(b) = f(ab^2)*f(a)$. Thus $\frac{f(a)}{f(b)} = \frac{f(a^2b)}{f(ab^2)}$ Combining these 2 we get $\frac{f(a^2c^2)}{f(b^2c^2)} = \frac{f(a^2b)}{f(ab^2)} \implies f(a^2c^2)*f(ab^2) = f(b^2c^2)*f(a^2b)$ $f(a^6b^4c^4) = f(a^4b^6c^4) \implies f(a)*f(a^2b^2c^2) = f(b)*f(a^2b^2c^2) \implies f(a)=f(b)$. This assumes that $f(a) = f(a^2b^2) \implies f(b) = 1$ which is a contradiction.

I think this solution works as well.. We denote $g(x)=f(2^x)$. The condition gives $g(a)g(b)=g(2a+2b)$ for all positive integers $a,b$. Therefore, we have $g(2a+4)=g(a)g(2)=g(a+1)g(1)$, for $a \ge 1$. We have $g(a+1)=g(a)*\frac{g(2)}{g(1)}$, so we have $g(n)=g(2)\cdot \frac{g(2)^{n-2}}{g(1)^{n-2}}$ for $n \ge 2$. Now let $\frac{g(2)}{g(1)} = k$. It is clear that $k$ is an integer. We have $g(n)=g(2) \cdot k^{n-2}$, so plugging this in, we have $g(2)^2k^{a+b-4} = g(2)k^{2a+2b-2}, g(2)=k^{a+b+2}$ for $a,b \ge 2$. Therefore, it is clear that $k=1$ and $g(2)=1$. We have $f(4)=1$, which is a contradiction. $\blacksquare$

Wlog pick ${a,b,c} \subset S$ so that $a>b>c$ . $\blacksquare f(a^8b^8c^8)=f(a^4)f(b^4c^4)=f(b^4)f(a^4c^4)\implies (f(a))^2f(b^2)f(c^2)=(f(b))^2f(a^2)f(c^2) \implies \left( \frac{f(a)}{f(b)} \right)^2=\frac{f(a^2)}{f(b^2)}$ $\blacksquare f(a^4b^4c^4)=f(a^2)f(b^2c^2)=f(b^2)f(a^2c^2) \implies f(a^2)f(b)f(c)=f(b^2)f(a)f(c) \implies \frac{f(a^2)}{f(b^2)} = \frac{f(a)}{f(b)}$ So , we infer that $f(a)=f(b)$ for all $a>b\geq3$ . Now , things are evident - we can easily derive a contradiction .So no such function.

samrocksnature wrote: @above what was ur motivation for g(x) the new function becomes nicer and easier to work with (it's also a common trick to take logs of these types of functions)

samrocksnature wrote: @above what was ur motivation for g(x) Changes the multiplicative form into an additive one.

Claim $: f(ab)f(cd) = f(ac)f(bd)$. Proof $: f(ab)f(cd) = f(a^2b^2c^2d^2) = f(ac)f(bd)$. So we have $f(2a)f(b) = f(2b)f(a) \implies \frac{f(2a)}{f(a)} = \frac{f(2b)}{f(b)}$. Also we have $f(4a^2)f(b) = f(2)f(a)f(b) = f(4b^2)f(a) \implies f(2ab)f(2a) = f(2ab)f(2b) \implies f(2a) = f(2b)$. So now we have $\frac{f(2a)}{f(a)} = \frac{f(2b)}{f(b)} , f(2a) = f(2b) \implies f(a) = f(b) \implies f(a) = c$ Putting that in main equation we have $c^2 = c \implies c = 0$ or $1$ which both are out of $S$ so there exists no such $f$.

I don't know if this is correct, but if it is then this is too easy for an APMO Let $a=2, b=3$ and let's take some $c \neq 2,3,36$. Then: $f(\frac{(abc)^4}{a^2})=f(a^2(b^2c^2)^2)=f(a)f(b)f(c)=f(\frac{(abc)^4}{b^2})=f(\frac{(abc)^4}{c^2}).$ But also: $f(\frac{(abc)^4}{a^2})f(a^2)=f((abc)^8)$ (Clearly $a^2 < \frac{(abc)^4}{a^2}$). Thus: $\frac{f(abc)^8}{f(a^2)}=\frac{f(abc)^8}{f(b^2)}=\frac{f(abc)^8}{f(c^2)}$ Which implies $k=f(4)=f(a^2)$ for all $a\neq 36$. But then: $kf(3)=f(4)f(3)=f(12^2)=k$, which gives us $k=0$ or $f(3)=1$, both impossible.

Suppose such an $f$ existed. Define the function $g: S \cup \{1\} \rightarrow S$ such that $g(x) = f(2x)$ for all $x \geq 1$. The condition becomes $g(a)g(b) = g(2a^2b^2)$ for all $a \neq b$. Then, for all $a > 8$ we have: \[g(2^5a^4) = g(2a^2)g(2) = g(a)g(1)g(2) = g(a)g(2^3) = g(2^7a^2).\] Plugging in $a = 2^5, 2^9, 2^{17}$, we obtain: \[g(2^{73}) = g(2^{41}) = g(2^{25}) = g(2^{17})\]. However, from our original condition we have $g(2^{17})g(2^{19}) = g(2^{73}) = g(2^{17})$ and so $g(2^{19}) = 1 \notin S$. Contradiction. Hence no such function exists.

Claim: $f(2^k) = f(2^{k+1}) = f( 2^{k+2})$. proof: We can see that \[f(2^k) \cdot f(2^{k+1}) \cdot f( 2^{k+2}) = f(2^{4k+2}) \cdot f(2^{k+2}) = f(2^{10k+8}) \]\[f(2^k) \cdot f(2^{k+1}) \cdot f(2^{k+2}) = f(2^{4k+4}) \cdot f(2^{k+1}) = f(2^{10k+10}) \]\[f(2^k) \cdot f(2^{k+1}) \cdot f(2^{k+2}) = f(2^k) \cdot f(2^{4k+6}) =f(2^{10k+12})\]Thus we can see that $f(2^{10k+8}) = f(2^{10k+10}) = f(2^{10k + 12})$. We can see that \[f(2^{10k+8}) = f(2^{4k+4}) \cdot f(2^k) = f(2^{10k+10}) = f(2^{4k+4}) \cdot f(2^{k+1}) = f(2^{10k+12}) = f(2^{4k+4}) \cdot f(2^{k+2}\] We can thus obviously see that $f(2^x) = c$ for some constant $c$. We know that $f(2) \cdot f(4) = f(64)$, or $c^2 = c$. This implies that $c \not \in S$, so there are no such functions. $\blacksquare$

Somehow different solution? Fix a prime $p$ and let $g(n) = \nu_p(g(n))$. Then we'll prove that $g(n) = 0$ for all $n \in S$, solving the problem. The condition can we rewritten as $g(a) + g(b) = g(a^2b^2)$. Consider the following claim: Claim: $n \cdot g(a) = g(a^{3 \cdot 2^{n-1} - 2})$ for all $n$. Proof. Induction on $n$ implies it. $\blacksquare$ Now let $m, n$ be a large integers. Then $g(a^{3 \cdot 2^{n+m-1}-2}) = (n + m)g(a) = g(a^{3 \cdot 2^{n-1}-2}) + g(a^{3 \cdot 2^{m-1}-2}) = g(a^{3(2^n + 2^m) - 4})$. On the other hand, we have $g(a^{3 \cdot 2^{n+m-1}-2}) = g(a^{3(2^n + 2^m) - 4}) + g(a^{6+3(2^{n+m-1}-2^{n+1}-2^{m+1})})$, so $g(a^{6+3(2^{n+m-1}-2^{n+1}-2^{m+1})}) = 0$ for all $a \in S$. Thus there exists large $N$ such that $g(a^N) = 0$, so $g(a^{4N}) = 2g(a^N) = 0$ and $0 = g(a^{4N}) = g(a) + g(a^{2N-1})$, therefore $g(a) = 0$ for all $a$. This completes proof. $\blacksquare$

Answer is no. Look exclusively at powers of $2$. Let $g(k) = f(2^{2k})$. We have $f(a^4b^4c^2) = (f(a)f(b))f(c) = (f(a)f(c))f(b) = f(a^4b^2c^4)$. Thus converting to $g$, we have: $g(2a + 2b + c) = g(2a + b + 2c)$ for all positive $a, b, c$. This is sufficient to show that $g(k)$ is constant for large $k$. However this means: $f(4^{\infty})f(4^{\infty}) = f(4^{\infty})$, giving contradiction as desired.

Note that $f(a)f(bk) = f(k^2a^2b^2) = f(ak)f(b) \implies \frac{f(a)}{f(b)} = \frac{f(ak)}{f(bk)}$. Moreover, we see that by setting $a=b=x$ we have $f(x)^2 = f(x^4)$ Thus, we do the following calculations \[ \frac{f(a)}{f(b)} = \frac{f(a)f(2)}{f(b)f(2)} = \frac{f(4a^2)}{f(4b^2)} = \frac{f(a^2)}{f(b^2)} = \frac{f(a^4)}{f(b^4)} = \left ( \frac{f(a)}{f(b)} \right )^2 \] So, $f(a)=f(b)=c$ for all $a$ and $b$ and some fixed $c$. So, we have $c^2 = f(a)f(b) = f(a^2b^2) = c \implies c=1$ which contradicts $1 \not \in S$.

We claim that no such functions exist. Let $a,b,c$ be distinct numbers from $S$. We have, \[f(a^4b^4c^2)=f(a^2b^2)f(c)=f(a)f(b)f(c)=f(a^4b^4c^2)=f(a)f(ab^2c)\]\[f(a^2b^4c^4)=f(b^2c^2)f(a)=f(a)f(b)f(c)=f(b^4c^4a^2)=f(c)f(ab^2c)\]So, $f(a)f(ab^2c)=f(c)f(ab^2c)$ which forces $f(a)=f(c)$, so $f$ is constant, say $K$. But then $K^2=K$, which means $K=0,1\notin S$, contradiction.

Note that \(f(36) = f(2) f(3) > f(2)\), and if \(ab = cd\) where \(a \neq b, c \neq d\), then \[f(a) f(b) = f(a^2 b^2) = f(c^2 d^2) = f(c) f(d)\text{.}\]So we have \begin{align*} f(36) f(4) &= f(2) f(72) = f(144^2) \\ f(36) f\left(\frac{144^2}{18}\right) &= f(2) f(144^2) = f(2) f(36) f(4) \\ f\left(\frac{144^2}{18}\right) &= f(2) f(4) = f(8^2). \end{align*}However, $f(2) f\left(\frac{144^2}{18}\right) = f(36) f(8^2)$ too, so $f(2) = f(36)$, contradiction. Therefore, no such \(f\) exists.

we have $f(a^7)f(a)=f(a^{16})=f(a^2)f(a^6)=f(a^2)^2f(a)=f(a^8)f(a)$ thus $f(a^7)=f(a^8)$ but then for any $b$ $$f(ab)f(a^7)=f(a^{16}b^2)=f(b)f(a^8)$$thus $f(ab)=f(b)$, since $a,b$ where arbitrary this gives that $f(x)=f(y)$ whenever $x|y$ but then $f(a)=f(a^4)=f(a)^2\implies f(a)=1$. Therefore, there are no such functions.

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Attachments:

Lemma : $f(k^2)f(t^2)=f(k^2t)f(t)$ Proof : $f(k^2)f(t^2)=f(k^4t^4)=f((k^2t)^2t^2)=f(k^2t)f(t)$ Thus $f(a^4)f(b^4)=f(a^4b^2)f(b^2)=f(a^2)f(b)f(b^2)$. By symmetry $f(a)=f(b)$ thus $f$ constant which can't be the case.

Notice that for any positive integers $a, b$ such that $a \neq 2b$ or vice versa, we have $f(a)f(2b) = f(2a)f(b)$, i.e. $\frac{f(2a)}{f(a)} = \frac{f(2b)}{f(b)}$. So in particular, $c = \frac{f(2n)}{f(n)}$ is constant for all $n \geq 2$. [We can clearly circumvent the $a \neq 2b$ condition.] But then \begin{align*} c^5 f(2) &= f(64) = f(2)f(4) = cf(2)^2 \\ c^7 f(2) &= f(256) = f(2)f(8) = c^2f(2)^2 \end{align*}The two equations imply $c^4 = f(2) = c^5$, so $c=1$ and $f(2) = 1$, contradiction.