For $n in mathbb{Z}^+$, $x in mathbb{Z}$, $k in mathbb{Z}$, $F(n)=min(|n^2-17x|)$ must be $0$ or $2^k$












0












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I've difficult to understand the question that was given to me yesterday:




Prove that for $n in mathbb{Z}^+$, $x in mathbb{Z}$, $k in mathbb{Z}$,
$F(n)=min(|n^2-17x|)$ must be $0$ or $2^k$




I've divided into two cases: if $n = 2i+1$ (odd) and $n = 2i$ (even), but I'm stuck at this point.



Any hints? I do not understand the || in this context.



Thanks






elementary-number-theory







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    $begingroup$
    The $||$ just means absolute value (the quantity inside is an integer).
    $endgroup$
    – platty
    Dec 5 '18 at 7:42
















0












$begingroup$


I've difficult to understand the question that was given to me yesterday:




Prove that for $n in mathbb{Z}^+$, $x in mathbb{Z}$, $k in mathbb{Z}$,
$F(n)=min(|n^2-17x|)$ must be $0$ or $2^k$




I've divided into two cases: if $n = 2i+1$ (odd) and $n = 2i$ (even), but I'm stuck at this point.



Any hints? I do not understand the || in this context.



Thanks






elementary-number-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The $||$ just means absolute value (the quantity inside is an integer).
    $endgroup$
    – platty
    Dec 5 '18 at 7:42














0












0








0





$begingroup$


I've difficult to understand the question that was given to me yesterday:




Prove that for $n in mathbb{Z}^+$, $x in mathbb{Z}$, $k in mathbb{Z}$,
$F(n)=min(|n^2-17x|)$ must be $0$ or $2^k$




I've divided into two cases: if $n = 2i+1$ (odd) and $n = 2i$ (even), but I'm stuck at this point.



Any hints? I do not understand the || in this context.



Thanks






elementary-number-theory







share|cite|improve this question









$endgroup$




I've difficult to understand the question that was given to me yesterday:




Prove that for $n in mathbb{Z}^+$, $x in mathbb{Z}$, $k in mathbb{Z}$,
$F(n)=min(|n^2-17x|)$ must be $0$ or $2^k$




I've divided into two cases: if $n = 2i+1$ (odd) and $n = 2i$ (even), but I'm stuck at this point.



Any hints? I do not understand the || in this context.



Thanks






elementary-number-theory




elementary-number-theory






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share|cite|improve this question











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asked Dec 5 '18 at 7:25









AlessarAlessar

20613




20613








  • 1




    $begingroup$
    The $||$ just means absolute value (the quantity inside is an integer).
    $endgroup$
    – platty
    Dec 5 '18 at 7:42














  • 1




    $begingroup$
    The $||$ just means absolute value (the quantity inside is an integer).
    $endgroup$
    – platty
    Dec 5 '18 at 7:42








1




1




$begingroup$
The $||$ just means absolute value (the quantity inside is an integer).
$endgroup$
– platty
Dec 5 '18 at 7:42




$begingroup$
The $||$ just means absolute value (the quantity inside is an integer).
$endgroup$
– platty
Dec 5 '18 at 7:42










1 Answer
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$begingroup$

You should instead divide into $17$ cases. The quantity $F(n)$ is entirely determined by the remainder of $n$ when divided by $17$. To prove this, show that if $n$ and $m$ have the same remainder upon division by $17$, then so do $n^2$ and $m^2$, and realize that the remainders of $n^2$ and $m^2$ determine their distance to the nearest multiple of $17$.



So you only need look at $n=1,2,dots,17$, and note that for each, $F(n)=0,1,2,4$ or $8$.






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    $begingroup$

    You should instead divide into $17$ cases. The quantity $F(n)$ is entirely determined by the remainder of $n$ when divided by $17$. To prove this, show that if $n$ and $m$ have the same remainder upon division by $17$, then so do $n^2$ and $m^2$, and realize that the remainders of $n^2$ and $m^2$ determine their distance to the nearest multiple of $17$.



    So you only need look at $n=1,2,dots,17$, and note that for each, $F(n)=0,1,2,4$ or $8$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You should instead divide into $17$ cases. The quantity $F(n)$ is entirely determined by the remainder of $n$ when divided by $17$. To prove this, show that if $n$ and $m$ have the same remainder upon division by $17$, then so do $n^2$ and $m^2$, and realize that the remainders of $n^2$ and $m^2$ determine their distance to the nearest multiple of $17$.



      So you only need look at $n=1,2,dots,17$, and note that for each, $F(n)=0,1,2,4$ or $8$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You should instead divide into $17$ cases. The quantity $F(n)$ is entirely determined by the remainder of $n$ when divided by $17$. To prove this, show that if $n$ and $m$ have the same remainder upon division by $17$, then so do $n^2$ and $m^2$, and realize that the remainders of $n^2$ and $m^2$ determine their distance to the nearest multiple of $17$.



        So you only need look at $n=1,2,dots,17$, and note that for each, $F(n)=0,1,2,4$ or $8$.






        share|cite|improve this answer









        $endgroup$



        You should instead divide into $17$ cases. The quantity $F(n)$ is entirely determined by the remainder of $n$ when divided by $17$. To prove this, show that if $n$ and $m$ have the same remainder upon division by $17$, then so do $n^2$ and $m^2$, and realize that the remainders of $n^2$ and $m^2$ determine their distance to the nearest multiple of $17$.



        So you only need look at $n=1,2,dots,17$, and note that for each, $F(n)=0,1,2,4$ or $8$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 7:44









        Mike EarnestMike Earnest

        20.9k11951




        20.9k11951






























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