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Find all functions $ f $ taking real numbers to positive integers, such that $$ f ^ { f ( x ) } ( y ) = f ( x ) f ( y ) $$ holds true for all real numbers $ x $ and $ y $.

Here $ f ^ n ( x ) $ is the $ n $-th iteration of $ f $ applied to $ x $; i.e. $ f ^ 0 ( x ) = x $ and $ f ^ { n + 1 } ( x ) = f \big( f ^ n ( x ) \big) $.

I think I can use $ f ( 0 ) = x $ but that just gets me stuck, as you can plug in $ x = 0 $ and $ y = 0 $ to get $ x ^ 2 = f ^ x ( 0 ) $, which I don't know how to solve. Any ways someone recommends to start this off?

Mohsen Shahriari
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1 Answers1

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It is clear that the constant function $f = 1$ satisfies the equation.

We show that there is no other possibility, by showing that the function $f$ cannot take any value greater than $1$.

Suppose that there exists $a$ such that $f(a) = k > 1$.

We put $x = a$ in the original equation: $$f^k(y) = kf(y)\tag{1}$$ for all $y$.

Put $y = a$ in (1): $f^{k - 1}(k) = k^2$.

Put $y = f^{k - 2}(k)$ in (1): $f^{2k - 2}(k) = kf^{k - 1}(k) = k^3$.

Put $y = f^{2k - 3}(k)$ in (1): $f^{3k - 3}(k) = kf^{2k - 2}(k) = k^4$.

Etc. This procedure continues, and we may show by induction that $$f^{m(k - 1)}(k) = k^{m + 1} \tag{2}$$ for all integer $m \geq 1$.

Now put $x = f^{k - 2}(k)$ in the original equation: $$f^{k^2}(y) = k^2 f(y)\tag{3}$$ for all $y$.

Put $y = a$ in (3): $f^{k^2 - 1}(k) = k^3$.

However, putting $m = k + 1$ in (2) gives $f^{k^2 - 1}(k) = k^{k + 2} > k^3$. This is a contradiction.

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