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Math Help - Find function

  1. #1
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    Find function

    I have to find functions f :  \mathbb{N} \mapsto  \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.
    Last edited by veileen; April 19th 2011 at 03:56 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by veileen View Post
    I have to find functions f :  \mathbb{N} \mapsto  \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.
    Last edited by topsquark; April 18th 2011 at 02:36 PM.
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  3. #3
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    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 !
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  4. #4
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    "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.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by veileen View Post
    "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
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  6. #6
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    Mhm, it's just x=0; sorry.
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