$f(xy) = f(x+y)$ : A variation of an old problem

Question

I have seen the following problem in various places, such as Toppr and Quora.

Suppose that the function $f:\mathbb{N}\to\mathbb{N}$ satisfies $f(x+y)=f(xy)$ for all $x,y\in\mathbb{N}$. Show that $f$ is constant.

This is an easy problem, because all we have to do is set $y=1$. But here's a variant that occurred to me. Suppose that the functional equation $f(x+y)=f(xy)$ is only required to hold for all integers $x\ge 2$ and $y\ge 2$. Does it follow that $f(x)$ is constant for all $x\ge 5$?

If we pick numbers $x_0$ and $y$ with $y\ge 2$ and $x_0-y\ge 2$, then $$f(x_0) = f((x_0-y) + y) = f((x_0-y)y),$$ so we can define $x_{n+1} := (x_n-y)y$ for $n\ge 0$ and conclude that $f$ is constant on the sequence $x_0,x_1,x_2,\ldots\,$. Picking various values of $(x_0,y)$, we find that $f$ is constant on each of the following sequences, and hence $f(x)$ is constant for $5\le x\le 20$ at least:

$$ \eqalign{(5,2) &: 5,6,8,12,20,36,\ldots \cr (5,3) &: 5, 6, 9, 18, \ldots\cr (7,2) &: 7, 10, 16, 28, \ldots\cr (7,3) &: 7,12,27\ldots\cr (9,2) &: 9,14\ldots\cr (11,2) &: 11,18,32\ldots\cr (13,3) &: 13,30, \ldots\cr (13,4) &: 13,36, \ldots\cr (15,2) &: 15,26,48, \ldots\cr (15,3) &: 15,36, \ldots\cr (17,2) &: 17,30, \ldots\cr (19,3) &: 19,48, \ldots\cr } $$ But I don't know if $f(x)$ is constant for all $x\ge 5$.

Answer 1

For the initial case notice $$ f(5)=f(2+3)=f(2\cdot3)=f(6). $$

For $n \geq 6$ let $x=n-4$ (so $x\geq 2$) and write

$$ f(n)=f(x+4)\stackrel{*}{=}f(4x)\stackrel{*}{=}f(2x\cdot 2x)=f(4x\cdot x)\stackrel{*}{=}f(5x)\stackrel{*}{=}f(x+5)=f(n+1). $$ where $*$ denotes use of the functional equation.

Answer 2

First verify the conjecture by hand for all $n\le 97$.

Then assume it is false in general. As noted by lulu, the smallest counterexample must be a prime, $p>97$.

If $p\equiv 5\pmod 6$, then $$ f(p) = f\left(5(p-5)\right)=f\left(30\cdot\frac{p-5}6\right)=f\left(30+\frac{p-5}{6}\right), $$ which equals $f(5)$, because $5\le 30+\frac{p-5}6<p$. Therefore, we must have $p\equiv 1\pmod 6$. Hence, $$ f(p)=f\left(13(p-13)\right) = f\left(78\cdot\frac{p-13}6\right)=f\left(78+\frac{p-13}{6}\right), $$ which also equals $f(5)$, because $5\le 78+\frac{p-13}6<p$. We have reached a contradiction and are done.

Answer 3

Consider a number which can be factored in two seperate ways, say $abcd$ with factoriztions $(ab,cd)$ and $(ac,bd)$.

We can see that (for large enough $ab,cd,ac,bd$) we have $f(ab+cd)=f(ac+bd)$.

Suppose now that we fix $a=2,d=2+1=3$ then for any $b,c \geq1$, $f(2b+3c)=f(2c+3b)$.

Now let $b=n$ and $c=n+a$, we now have $f(5n+3a)=f(2n+3(n+a))=f(2(n+a)+3n)=f(5n+2a)$.

Now consider the following collection of equalities:

$$5(n+2)=2(n+5)+3n$$ $$5(n+3)=2n+3(n+5)$$
$$5n+4=2(n+2)+3n$$ $$5(n+1)+1=2n+3(n+2)$$
$$5(n+1)+1=2(n+3)+3n$$ $$5(n+1)+4=2n+3(n+3)$$
$$5(n+1)+2=2(n+2)+3(n+1)$$ $$5(n+1)+3=2(n+1)+3(n+2)$$
$$5(n+1)+3=2(n+4)+3n$$ $$5(n+2)+2=2n+3(n+4)$$

These together with the above observations tell us that for $x\geq 11$ the value of $f(x)$ depends only on the equivalence class of $x$ mod $5$.

As the original question already showed $f(x)$ is constant for $5 \leq x \leq 20$ we can conclude that $f(x)$ is constant for all $x \geq 5$. $\square$

Answer 4

To eliminate checking small cases, consider

$$f(N) = f(N-2+2) = f(2(N-2)) = f((N-2)+(N-2)) = f((N-2)^2)$$

We only needed $N \ge 4$ to assume $N-2 \ge 2$, but if we take $N \ge 5$ then we have $(N-2)^2 > N$ as well, so given any starting $N_0 \ge 5$, we have an increasing sequence $N_0 < N_1 < N_2 < \ldots$ on which $f$ is constant. Since it is a sequence of integers, it grows arbitrarily large. So as long as $f$ is constant for all sufficiently large $N$, it is constant for all $N$.

As suggested in the comments, suppose $f$ is not constant and let $p$ be the smallest prime such that $f(p) \ne f(5)$. Then since $p$ is odd,

$$f(p) = f(\tfrac{p-3}{2}+\tfrac{p-3}{2}+3) =f(3(\tfrac{p-3}{2})^2)=f((\tfrac{p-3}{2})^6)=f((\tfrac{p-3}{2}+6)$$

provided $\tfrac{p-3}{2} \ge 8$ (i.e. $p \ge 19$) for the final equality. In this case, $\tfrac{p-3}{2}+6 < p$, so $p$ was not a minimal counterexample. The cases $p<19$ are covered by the OP.