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I saw this problem on a mock test. Find all functions $f : \mathbb{N^+} \rightarrow \mathbb{N^+}$ such that $f(f(x))=9x$ holds for all positive integers $x$.

It is quite easy to see that $f(x)=3x$ satisfies the original conditions (and I suspect that it is the only answer), however, I can't even prove that $f(x)$ is linear (which will derive that $f(x)=3x$, since it is operating on $\mathbb{N^+}$).

My only progress on the problem was that $f(x)$ is injective:

If $f(x)=f(y)$ then

$f(f(x))=f(f(y))$

Using the original equation gives:

$x=y$

Thomas Andrews
  • 186,215
RandomGuy
  • 105

2 Answers2

1

Ok. So I solved it, thanks to @lulu's comment, which made me realize that $f(x)=3x$ might NOT be the only answer.

After realizing that and getting out of the trapped box, the answer was rather straightforward. Note that the only REAL constraint is that $f(9^xy)=9^xf(y)$.

I found two types of answers satisfying this property:

Type $1$:

First, define $S = \{x \mid x \in \mathbb{N^+}, 9 \nmid x\}$. Then for each ordered tuple $(U, V, g)$ such that the following holds:

  • $U \cup V = S$
  • $U \cap V = \emptyset$
  • $g : U \rightarrow V$
  • $g$ is bijective (aka $g^{-1} : V \rightarrow U$)

Then we have a corresponding $f$, defined as follows:

$f = \begin{cases} 9^u * f(v) & 9 \mid x, x = 9^u \times v \\ g(x) & x \in U \\ 9 \times g^{-1}(x) & x \in V \end{cases}$

Proof that it works:

If $x \in U$ then $f(f()x) = f(g(x)) = 9 \times g^{-1}(g(x)) = 9x$ (since $g(x) \in V$).

If $x \in V$ then $f(f(x)) = f(9 \times g^{-1}(x)) = 9 \times f(g^{-1}(x)) = 9 \times g(g^{-1}(x)) = 9x$ (since $g^{-1}(x) \in U$).

If $9 \mid x$ then after $x = 9^u \times v$, $f(f(x)) = f(9^u f(v)) = 9^u f(f(v)) = 9^u \times 9v = 9x$

(In the last case, $f(f(v))=9v$ is proved in the first two cases).

Type $2$:

This is the case where $\forall x, 3 \mid f(x)$.

First, define $S = \{x \mid x \in \mathbb{N^+}, 3 \nmid x\}$. Then for each ordered tuple $(U, V, g)$ such that the following holds:

  • $U \cup V = S$
  • $U \cap V = \emptyset$
  • $g : U \rightarrow V$
  • $g$ is bijective (aka $g^{-1} : V \rightarrow U$)

Then we have a corresponding $f$, defined as follows:

$f = \begin{cases} 9^u * f(v) & 9 \mid x, x = 9^u * v \\ 9 \times g(\frac{x}{3}) & 3 \mid x, 9 \nmid x, \frac{x}{3} \in U \\ 9 \times g^{-1}(\frac{x}{3}) & 3 \mid x, 9 \nmid x, \frac{x}{3} \in V \\ 3 \times g(x) & x \in U \\ 3 \times g^{-1}(x) & x \in V \end{cases}$

I will leave the proof as an EXERCISE :).

End Notes:

For anyone of you who want an explicit function that satisfies $f(f(x))=9x$ but $f(x) \neq 3x$, here is one from Type $1$:

$f = \begin{cases} 9^u * f(v) & 9 \mid x, x = 9^u \times v \\ 9(x-1) & x \equiv 2, 4, 6, 8 \pmod 9 \\ x + 1 & x \equiv 1, 3, 5, 7 \pmod 9 \end{cases}$

RandomGuy
  • 105
-1

We are asked to find all functions $f: \mathbb N \to\mathbb N$ such that:

$f(f(x))=9x$ for all $x \in \mathbb N$

for all $x \in \mathbb N$ ,define a tuple

$(x,f(x),f(f(x)),9x)$

We interpret this as a chain of constraint

choosing $f(x)$ forces $f(f(x))$,and

That must exactly be equal to $9x$

We interpret each 4-tuple $(x,f(x),f(f(x)),9x)$ as a structure in 2D or 3D space in particualr $(x,f(f(x)))$ = $(x,9x)$ form a line

{$(x,9x)| x \in \mathbb N$}

This is a straight line in the cartesian plane any deviation from the rule $f(x) = 3x$ would create a bend or discontinuity in the ordered structure.

PAPLOAF2008
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