Let $f$ be a function $f : \mathbb{N} \to \mathbb{N}$ such that $f(2x+3y)=f(x)f(y)$, determine $f$

Question

here what I did .

$$f(0)=f(0)^2$$ so $f(0)=1$ or $f(0)=0$ IF $f(0)=1$ we have $f(2y)=f(y)$ $$f(1)=f(2)=\ldots=f(2^n)=a$$ the equation $f(x)-a=0$ has infinitly many solutions , so $f(x)=a$

since f(0)=1 , $f(x)=1$.

i don't know how to handle the other case , i just found that $f(2x)=f(3x)=0$

Answer 1

A priori, there is no reason to expect $f$ to be a polynomial. So knowing that $f(x) = 0$ for infinitely many $x$ doesn't mean that $f$ is identically zero.

I consider the case when $f(0) = 1$. Then $f(2x + 0y) = f(x)$, so $f(x) = f(2x)$ for all $x$. Similarly, $f(x) = f(3x)$ for all $x$.

Let's evaluate $f(1)$.

Note that $f(5) = f(1\cdot 2 + 1\cdot 3) = f(1)^2$.

Also note that $13 = 5 \cdot 2 + 1 \cdot 3$ and $13 = 2 \cdot 2 + 3 \cdot 3$. Thus $f(13) = f(5)f(1) = f(1)^3$ and $f(13) = f(2)f(3) = f(1)^2$.

Thus $f(1)^2 = f(1)^3$, forcing $f(1) = 0$ or $f(1) = 1$.

It is a straightforward induction argument to show that $f(n)$ is a power of $f(1)$ for all $n \geq 1$. In either case, it is now apparent that $f(0) = 1$ and $f(n) = 1$ or $f(n) = 0$ for $n \geq 1$ are two possible solutions.

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If $f(0) = 0$, then $f(2x + 0y) = f(x)f(0) = 0$, and similarly $f(3x) = 0$ for all $x$.

Let's evaluate $f(1)$ again.

Now note that $f(13) = f(5 \cdot 2 + 1 \cdot 3) = f(5)f(1)$, and $f(5) = f(1 \cdot 3 + 1 \cdot 2) = f(1)f(1)$, so taht $f(13) = f(1)^3$ (as above). But $f(13) = f(2 \cdot 2 + 3 \cdot 3) = f(2)f(3) = 0$, so that $f(1) = 0$.

Given any $n$ odd, we can write $n = 2x + 3y$ with $y = 1$. Then $f(n) = f(?)f(1) = 0$. Thus $f(n) = 0$ if $n$ is odd.

Given any $n$ even, as $f(2x) = 0$ for all $x$, we see that $f(n) = 0$. Thus we have shown that $f(x)$ is identically $0$ if $f(0) = 0$.

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Thus in total, there are three such functions. They are

• $f(n) = 0$ for all $n$.
• $f(n) = 1$ for all $n$.
• $f(0) = 1$, and $f(n) = 0$ for all $n \geq 1$.

Answer 2

If $$ f\left(2x\right)=f\left(3y\right)=0 \Rightarrow f\left(2x+3y\right)=0 $$

Let's take a natural $n$, then can be written as $n=2x_1+3x_2$ because

first case : $n$ is even, so $n=2x_1$

second case : $n$ is odd, so $n-3$ is even so $n=2x_1+3$

So for all $n$, $n=2x_1+3x_2$ hence $$ f=0 $$

EDIT $$ f\left(2x\right)=0=f\left(3y\right) \Rightarrow f\left(2x\right)f\left(3y\right)=0 \Rightarrow f\left(2x+3y\right)=0 $$ because of the definition of $f$

Answer 3

This is related to the Frobenius problem (or coins or Mcnuggets problem) of two numbers:

If $a,b\in\mathbb{N}$ are two numbers such that $\gcd(a,b)=1$, then all numbers $n\geq(a-1)(b-1)$ are representable as a nonnegative combination of $a,b$, i.e., $n=ax+by$ with $x,y\in\mathbb{N}$.

1) If $f(0)=0$:

Since $\gcd(4,3)=1$ and $(4-1)(3-1)=2$, if $n\geq6$, then there are $x_n,y_n\in\mathbb{N}$ such that $n=4x_n+3y_n$.

Since $f(0)=0$ we get $f(2x)=f(2x+3\cdot 0)=f(x)f(0)=0$ for every $x\in\mathbb{N}$, so $f(n)=f(4x_n+3y_n)=f(2x_n)f(y_n)=0$ for all $n\geq6$. Since $f(0)=f(2)=f(4)=0$, it just remains to show what happens with $f(1),f(3),f(5)$.

Note that $f(3)^2=f(3)f(3)=f(2\cdot 3+3\cdot 3)=f(15)=0$ (because $15\geq6$), so $f(3)=0$. Similarly, $f(5)^2=f(25)=0$, and finally $f(1)^2=f(5)=0$, so $f(1)=f(3)=f(5)=0$ and $f(x)=0$ for every $x\in\mathbb{N}$.

b) If $f(0)=1$:

Since $\gcd(2,3)=1$, for every $n\geq2$ there are $x_n,y_n\in\mathbb{N}$ such that $n=2x_n+3y_n$. Denote $a:=f(1)$. By strong induction, suppose that for every number $m<n$, $f(m)$ is some power of $a$ (we have as base cases $f(0)=a^0$, $f(1)=a^1$). Then $f(n)=f(2x_n+3y_n)$ with $0\leq x_n,y_n<n$, thus $f(n)=f(x_n)f(y_n)=a^ra^s=a^{r+s}$ for some exponents $r,s$.Note that we can do this in such a way that the exponent is never $0$ (except for $f(0)$).

On the other hand, the first number representable in two ways (with different numbers unregarding the order) as a nonnegative linear combination of $2$ and $3$ is $8=2\cdot 4+3\cdot 0=2\cdot 1+3\cdot 2$; thus $f(8)=f(4)f(0)=f(4)$ and $f(8)=f(1)f(2)$, so $f(4)=f(1)f(2)=af(2)$. But $f(4)=f(2\cdot 2 +3\cdot 0)=f(2)f(0)=f(2)$, hence $af(2)=f(2)$, so either $a=1$ or $f(2)=0$; and $f(2)=f(2\cdot 1+3\cdot 0)=f(1)=a$, so either $a=1$ or $a=0$. Noting that every other image is a power of $a$, we get either $f(x)=1$ for every $x$ or $f(0)=1$, $f(x)=0$ for every $x>0$. Observe that these functions indeed satisfy the given condition.