1. ## Find function

I have to find functions f : $\displaystyle \mathbb{N}$ $\displaystyle \mapsto$ $\displaystyle \mathbb{N}$ such that f(xf(y) + y) = yf(x) + f(y) (for all x, y natural numbers) and f(x) is a prime number for any prime number x.

Well... I don't have many ideas.

x=0: f(0) = 0
x=y: f(x(f(x)+1)) = f(x)(x+1), for any natural x.

I want to prove that f(x) = x.

2. Originally Posted by veileen
I have to find functions f : $\displaystyle \mathbb{N}$ $\displaystyle \mapsto$ $\displaystyle \mathbb{N}$ such that f(xf(y) + y) = yf(x) + f(y) (for all x, y natural numbers) and f(x) is a prime number for any prime number x.

Well... I don't have many ideas.

x=y=0: f(0) = 0
x=y: f(x(f(x)+1)) = f(x)(x+1), for any natural x.

I want to prove that f(x) = x.
I'm afraid I can't help much, but I thought I'd point out that, if you haven't been given f(0) = 0 you can't assume that from x = y = 0...
f( 0*f(0) + 0 ) = 0*f(0) + f(0)

So f(0) = f(0), not necessarily f(0) = 0.

-Dan

Edit: Perhaps we can do this. Let y = 0. Then f( x*f(0) ) = f(0). Now, we can't say that f(cx) = c for all x because of the other relation you stated: f( x(f(x) + 1) ) = (x + 1)f(x) because c is not equal to (x + 1)c for all x. Thus c = 0 is the only solution making f(0) = 0.

3. Well , actually f(0)=0
this is because when we let x=0 we get for every natural y f(y)=yf(0)+f(y) which is equivalent to y=0 (impossible for every natural y ) or f(0)=0 which is what we need .
apparently f:x|--->x is the only solution , and since we are in N , we HAVE to do it using induction !

4. "I'm afraid I can't help much, but I thought I'd point out that, if you haven't been given f(0) = 0 you can't assume that from x = y = 0..."
Uhm, I can... f(xf(y) + y) = yf(x) + f(y) (for all x, y natural numbers)

"we HAVE to do it using induction" Yup, but isn't it a bit early to do that? I know just that f(0)=0, nothing about f(x) in general. Even if I'm sure f(x)=x is the only solution, I have to prove this.

5. Originally Posted by veileen
"I'm afraid I can't help much, but I thought I'd point out that, if you haven't been given f(0) = 0 you can't assume that from x = y = 0..."
Uhm, I can... f(xf(y) + y) = yf(x) + f(y) (for all x, y natural numbers)
I'm not arguing that f(0) isn't 0, just that by plugging x = y = 0 doesn't give you f(0) = 0. It gives f(0) = f(0).

-Dan

6. Mhm, it's just x=0; sorry.