Let $f$ be a function $f : \mathbb{N} \to \mathbb{N}$ such that $f(2x+3y)=f(x)f(y)$, determine $f$ 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 $f(2x)=f(3x)=0$

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.

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 $f(0) = 0$.

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 $n \geq 1$.

