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Thread: Strong function

  1. #1
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    Strong function

    if so that

    1. prove that is not constant
    2. find
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  2. #2
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    Quote Originally Posted by dapore View Post
    if so that









    1. prove that is not constant

    2. find

    First of all, I'm guessing you intended to write $\displaystyle f:\mathbb{Z}\rightarrow \mathbb{Z}=$ a function from the integers to the integers.

    Second, if $\displaystyle \forall x\in\mathbb{Z}\,,\,\,f(x)=k=$ a constant, then from the given functional relation $\displaystyle f(x+y)+f(xy-1)=f(x)f(y)-2$ it

    follows that $\displaystyle 2k=k^2-2\Longrightarrow k^2-2k-2=0$ , and this quadratic has two real non-integer solutions, contradiction.

    Next, inputting

    (a) $\displaystyle x=y=0$ in the func. relation we get $\displaystyle f(0)+f(-1)=f(0)^2-2$ ;

    (b) inputting $\displaystyle x=y=-1$ we get $\displaystyle f(-2)+f(0)=f(-1)^2-2$ , and;

    (c) inputting $\displaystyle x=-1\,,\,y=0$ we get $\displaystyle 2f(-1)=f(-1)f(0)-2$

    Now you can find $\displaystyle f(-1)$ as a function of $\displaystyle f(0)$ and substitute then in (a) to get $\displaystyle f(0)$ and etc. (Do you know/remember how to search for rational/integer roots of integral polynomials? This can come handy in this case)
    Hmmm... the problem is that the only two possible values for $\displaystyle f(0)$ yield non-integer values for $\displaystyle f(-1)$... .Check this.

    I miscalculated: there's only one possible value for $\displaystyle f(0)$ which then gives one integer value for $\displaystyle f(-1)$ and then one unique integer value for $\displaystyle f(-2)$...fine!

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    First of all, I'm guessing you intended to write $\displaystyle f:\mathbb{Z}\rightarrow \mathbb{Z}=$ a function from the integers to the integers.

    Second, if $\displaystyle \forall x\in\mathbb{Z}\,,\,\,f(x)=k=$ a constant, then from the given functional relation $\displaystyle f(x+y)+f(xy-1)=f(x)f(y)-2$ it

    follows that $\displaystyle 2k=k^2-2\Longrightarrow k^2-2k-2=0$ , and this quadratic has two real non-integer solutions, contradiction.

    Next, inputting

    (a) $\displaystyle x=y=0$ in the func. relation we get $\displaystyle f(0)+f(-1)=f(0)^2-2$ ;

    (b) inputting $\displaystyle x=y=-1$ we get $\displaystyle f(-2)+f(0)=f(-1)^2-2$ , and;

    (c) inputting $\displaystyle x=-1\,,\,y=0$ we get $\displaystyle 2f(-1)=f(-1)f(0)-2$

    Now you can find $\displaystyle f(-1)$ as a function of $\displaystyle f(0)$ and substitute then in (a) to get $\displaystyle f(0)$ and etc. (Do you know/remember how to search for rational/integer roots of integral polynomials? This can come handy in this case)
    Hmmm... the problem is that the only two possible values for $\displaystyle f(0)$ yield non-integer values for $\displaystyle f(-1)$... .Check this.

    I miscalculated: there's only one possible value for $\displaystyle f(0)$ which then gives one integer value for $\displaystyle f(-1)$ and then one unique integer value for $\displaystyle f(-2)$...fine!

    Tonio


    1. assume the contrary, let for an integer and for all . Thus,


    which obviously has no solution for . Contradiction.

    2. Let be the assertion


    Let ,


    Plugging this information to (1),




    is an integer, thus . Consequently,
    And additionally,

    .

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