$f(3x+2y)=f(x)f(y)$ for all $x,y \in \mathbb{N}.$

Question

An idea to solve $f(3x+2y)=f(x)f(y)$ $\forall x,y \in \mathbb{N}$ ? I found function $f(0)=0$, is it great ? But I think, there is another function which works...

Answer 1

I assume that $0$ is not in the domain, so unlike other posts I won't be considering $f(0)$. My conclusion is the same though: either $f(n)=0$ for all $n$, or $f(n)=1$ for all $n$.

Case 1: Suppose $f(1)=0$.

For any odd $n\geq5$, you can write $n=3\cdot1+2y$, and then $f(n)=f(1)f(y)=0$. What about $f(3)$? Well, $0=f(15)=f(3\cdot3+2\cdot3)=f(3)^2$, so $f(3)$ must be $0$ as well. This establishes $f(n)=0$ for all odd $n$.

Then $f(4)f(2)=f(3\cdot4+2\cdot2)=f(16)=f(3\cdot2+2\cdot5)=f(2)f(5)=0$. So either $f(2)=0$ or $f(4)=0$.

Now let $n=2k>4$. $$\begin{align} f(n)^{2^k} &=\left(f(n)f(n)\right)^{2^{k-1}}\\ &=f(3n+2n)^{2^{k-1}}\\ &=f(3(n-2)+2(n+3))^{2^{k-1}}\\ &=f(n-2)^{2^{k-1}}f(n+3)^{2^{k-1}}&\text{note first argument is next smallest even number}\\ &=\cdots&\text{in exchange for reducing exponent power by one}\\ &=f(4)\cdot A\\ &=f(2)\cdot B \end{align} $$

Since either $f(4)$ or $f(2)$ is $0$, then so is $f(n)$. This establishes $f(n)=0$ for all even $n$ that are at least $4$.

Lastly $$f(2)^2=f(3\cdot2+2\cdot2)=f(10)=0$$ so $f(2)=0$ as well.

This establishes $f(n)=0$ for all $n$ in the case where $f(1)=0$.

Case 2: Suppose $f(1)\neq0$.

First, $$ \begin{align} f(1)^6 &=(f(1)f(1)f(1))^2 =(f(1)f(3\cdot1+2\cdot1))^2 =(f(1)f(5))^2\\ &=f(3\cdot1+2\cdot5)^2 =f(13)^2 =f(3\cdot3+2\cdot2)^2\\ &=f(3)^2f(2)^2 =f(3\cdot3+2\cdot3)f(2)^2 =f(15)f(2)^2\\ &=f(3\cdot1+2\cdot6)f(2)^2\\ &=f(1)f(6)f(2)^2 \end{align} $$ This was all to establish that $f(6)\neq0$.

Now $f(1)f(4)=f(3\cdot1+2\cdot4)=f(11)=f(3\cdot3+2\cdot1)=f(3)f(1)$. So we conclude $f(4)=f(3)$.

Then $f(18)=f(3\cdot2+2\cdot6)=f(2)f(6)$. But also $f(18)=f(3\cdot4+2\cdot3)=f(4)f(3)$. Remember that $f(4)=f(3)$, so we conclude $f(6)f(2)=f(3)^2$.

Then $f(1)f(6)=f(3\cdot1+2\cdot6)=f(15)=f(3\cdot3+2\cdot3)=f(3)^2=f(6)f(2)$. We established $f(6)\neq0$, so we conclude $f(2)=f(1)$.

Now $f(1)^2=f(3\cdot1+2\cdot1)=f(5)$.

And $f(1)^3=f(1)f(5)=f(3\cdot1+2\cdot5)=f(13)=f(3\cdot3+2\cdot2)=f(3)f(2)=f(3)f(1)$. So we conclude that $f(3)=f(1)^2$.

Taking stock so far, we know $$(f(1),f(2),f(3),f(4),f(5))=\left(f(1),f(1),f(1)^2,f(1)^2,f(1)^2\right)$$

Now $f(1)^4=f(3)^2=f(3\cdot3+2\cdot3)=f(15)=f(3\cdot1+2\cdot6)=f(1)f(6)$. So we conclude that $f(6)=f(1)^3$.

And $f(7)=f(3\cdot1+2\cdot2)=f(1)f(2)=f(1)^2$.

And $f(8)=f(3\cdot2+2\cdot1)=f(2)f(1)=f(1)^2$.

And $f(1)^3=f(1)f(8)=f(3\cdot1+2\cdot8)=f(19)=f(3\cdot3+2\cdot5)=f(3)f(5)=f(1)^4$. Now we can [finally] conclude that $f(1)=1$.

So we've established that $f(n)=1$ for all $n\leq8$. Now inductively, starting at $n=9$, we can break $n$ down as $3x+2y$ with $x$ and $y$ so small that we know $f(x)f(y)=1$. An induction argument leads to $f(n)=1$ for all $n$.

Answer 2

More generally, suppose for some $a, b \ne 0$, $\forall x, y$, $f(ax+by) = f(x)f(y) $.

Setting $x = y = 0$, $f(0) = f^2(0)$ so $f(0) = 0$ or $f(0) = 1$.

If $f(0) = 0$ then $f(by) = 0$ for all $y$ so $f$ is zero everywhere.

If $f(0) = 1$ then, setting $y = 0$, $f(ax) = f(x)$, so, if $a \ne 1$, $f(a^nx) = f(x)$ for $n \in \mathbb{N}^{\pm}$. If $f$ is continuous at zero, then $f(x) = 1$ for all $x$.

Similarly if $b \ne 1$.

If $a = b = 1$ then $f(x+y) = f(x)f(y)$. If $f$ is continuous at zero then $f(x) = a^x$ where $a = f(1)$.

Don't know what to do if $f$ not continuous at zero, so I'll leave it at this.

Answer 3

We weren't told what the domain of $f$ was. Based on what was written in the post, I'm going to assume that $\Bbb{N}$ is the the set of non-negative integers. I'll also assume that we have a function $f:\Bbb{N}\to\Bbb{N}$, such that for all $x,y\in\Bbb{N}$,

$$f(3x+2y)=f(x)f(y).$$

I will show that one of the following is true:

(1) $f(x)=0$ for all $x\in\Bbb{N}$.

(2) $f(x)=1$ for all $x\in\Bbb{N}$.

(3) $f(0)=1$, and $f(x)=0$ for all $x\in\Bbb{N}$ with $x\ge1$.

In the first two cases, it is not hard to verify that $f$ satisfies the given functional equation. To show that $f$ satisfies the functional equation in the third case, note that for any $x,y\in\Bbb{N}$, $3x+2y=0$ iff $x=y=0$. Hence for any $x,y\in\Bbb{N}$, either $(x,y)=(0,0)$, in which case $f(3x+2y)=1=f(x)f(y)$, or $(x,y)\ne(0,0)$, in which case $f(3x+2y)=0=f(x)f(y)$.

We have found three distince functions that satisfy the given equation. It remains to show that these are the only three. Let $f:\Bbb{N}\to\Bbb{N}$ satisfy $f(3x+2y)=f(x)f(y)$, for all $x,y\in\Bbb{N}$. Letting $x=y=0$, we get that

$$f(0)=f(3\cdot0+2\cdot0)=f(0)f(0),$$

So $f(0)=f(0)^2$. Hence either $f(0)=0$ or $f(0)=1$. So we have two cases.

Case 1: Suppose $f(0)=0$. Letting $x=0$ we have that

$$f(2y)=f(3\cdot0+2y)=f(0)f(y)=0\cdot f(y)=0,$$

which shows that $f(n)=0$ whenever $n$ is even. Letting $y=0$ we have that

$$f(3x)=f(3x+2\cdot0)=f(x)f(0)=f(x)\cdot0=0,$$

which shows that $f(n)=0$ whenever $3$ divides $n$. Letting $y=2$ we have that

$$f(3x+4)=f(3x+2\cdot2)=f(x)f(2)=f(x)\cdot0=0,$$

which shows that $f(n)=0$ whenever $n\equiv1$ modulo $3$ and $n\ge4$. Letting $y=4$ we have that

$$f(3x+8)=f(3x+2\cdot4)=f(x)f(4)=f(x)\cdot0=0,$$

which shows that $f(n)=0$ whenever $n\equiv2$ modulo $3$ and $n\ge8$. At this point, we've shown that $f(n)=0$ for every $n$ except $1$ and $5$. Let $z=f(1)$. Then

$$f(5)=f(3\cdot1+2\cdot1)=f(1)f(1)=z^2.$$

It follows that

$$z^3=f(1)f(5)=f(3\cdot1+2\cdot5)=f(13)=0.$$

Hence $z=0$, and $f(x)=0$ for all $x\in\Bbb{N}$. $\quad\Box$

Case 2: Suppose that $f(0)=1$. Note that

$$f(3)=f(3\cdot1+2\cdot0)=f(1)f(0)=f(1).$$

So $f(3)=f(1)$. Also

$$f(9)=f(3\cdot3+2\cdot0)=f(3)f(0)=f(3)=f(1).$$

So $f(9)=f(1)$, but also

$$f(9)=f(3\cdot1+2\cdot3)=f(1)f(3)=f(1)^2.$$

So we also have that $f(9)=f(1)^2$. Since $f(1)=f(1)^2$, either $f(1)=0$ or $f(1)=1$. So we have two more cases.

Case 2.1 Suppose $f(0)=1$ and $f(1)=0$. We have that

$$f(2y)=f(3\cdot0+2y)=f(0)f(y)=f(y).$$

So $f(2y)=f(y)$ for all $y\in\Bbb{N}$. Hence $f(4)=f(2)=f(1)=0$. We also have that

$$f(3+2y)=f(3\cdot1+2y)=f(1)f(y)=0,$$

which shows that $f(n)=0$ for every odd $n$ with $n\ge3$, and

$$f(6+2y)=f(3\cdot2+2y)=f(2)f(y)=0,$$

which shows that $f(n)=0$ for every even $n$ with $n\ge6$. We have shown that $f(0)=1$ and $f(x)=0$ for all $x\in\Bbb{N}$ with $x\ge1$. $\quad\Box$

Case 2.2 Suppose $f(1)=f(0)=1$. Note that

$$f(2y)=f(3\cdot0+2y)=f(0)f(y)=f(y).$$

So $f(2y)=f(y)$ for all $y\in\Bbb{N}$. Hence $f(4)=f(2)=f(1)=1$. We also have that

$$f(3+2y)=f(3\cdot1+2y)=f(1)f(y)=f(y),$$

so $f(3+2y)=f(y)$ for all $y\in\Bbb{N}$. Since we've already shown that $f(n)=1$ for all $0\le n\le4$, it follows by induction that $f(n)=1$ for all $n\in\Bbb{N}$. $\quad\Box$