Representation of an integer / Euclidean algorithm












3












$begingroup$


Let $r in mathbb{N}$ be a natural number. Let
$$L geq 2(r-1)²$$
A paper (on quantum information theory, I'm not an expert in number theory or so...) I'm recently reading now says



"One can easily check that $L$ can be represented as
$$L = a(r-1)+br$$
where $a, b in mathbb{Z}$ are positive integers and
$$K = a + b$$
is even."



My first attempt was of course to apply something like the euclidean algorithm, which gives us
$$L = k(r-1)+m$$
where $k, m$ are positive integers, $m<r-1$ (I tried to apply euclidean algorithm for $r$ instead of $r-1$ at first, but this seems to be more useful...)



As we have
$$L = k(r-1) + m geq 2(r-1)²$$
with $m leq r-2$, we know
$$k+1-m-2 geq 2r(r-1) -(r-2)-2 geq 2r² -3r -1 geq (r-1)²$$
for all $rgeq 2$. (The case $r=1$ is not interesting.)



Therefore we can write
$$L = (k+1)(r-1)+m-r+1 = (k+1-m-2)(r-1)+(m+1)r$$
where now $(k+1-m-2), (m+1)$ are positive integers. But their sum equals
$$(k+1-m-2)+(m+1)=k$$
which of course does not have to be even.



Can somebody give me a hint how to manipulate this way of writing
$$L=a(r-1)+br$$
where $a+b$ is even?



This whole thing is a little weird, if you consider e.g. $r=5, L=33geq 32$, you could write
$$33=5r+2(r-1)$$
where $5+2$ is not even, but you also can write
$$33=r+7(r-1)$$
where $1+7$ is even (I don't know if an expert in stuff like this would consider this as weird, but I do...)



Thanks for any help :)






summation integers euclidean-algorithm







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let $r in mathbb{N}$ be a natural number. Let
    $$L geq 2(r-1)²$$
    A paper (on quantum information theory, I'm not an expert in number theory or so...) I'm recently reading now says



    "One can easily check that $L$ can be represented as
    $$L = a(r-1)+br$$
    where $a, b in mathbb{Z}$ are positive integers and
    $$K = a + b$$
    is even."



    My first attempt was of course to apply something like the euclidean algorithm, which gives us
    $$L = k(r-1)+m$$
    where $k, m$ are positive integers, $m<r-1$ (I tried to apply euclidean algorithm for $r$ instead of $r-1$ at first, but this seems to be more useful...)



    As we have
    $$L = k(r-1) + m geq 2(r-1)²$$
    with $m leq r-2$, we know
    $$k+1-m-2 geq 2r(r-1) -(r-2)-2 geq 2r² -3r -1 geq (r-1)²$$
    for all $rgeq 2$. (The case $r=1$ is not interesting.)



    Therefore we can write
    $$L = (k+1)(r-1)+m-r+1 = (k+1-m-2)(r-1)+(m+1)r$$
    where now $(k+1-m-2), (m+1)$ are positive integers. But their sum equals
    $$(k+1-m-2)+(m+1)=k$$
    which of course does not have to be even.



    Can somebody give me a hint how to manipulate this way of writing
    $$L=a(r-1)+br$$
    where $a+b$ is even?



    This whole thing is a little weird, if you consider e.g. $r=5, L=33geq 32$, you could write
    $$33=5r+2(r-1)$$
    where $5+2$ is not even, but you also can write
    $$33=r+7(r-1)$$
    where $1+7$ is even (I don't know if an expert in stuff like this would consider this as weird, but I do...)



    Thanks for any help :)






    summation integers euclidean-algorithm







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $r in mathbb{N}$ be a natural number. Let
      $$L geq 2(r-1)²$$
      A paper (on quantum information theory, I'm not an expert in number theory or so...) I'm recently reading now says



      "One can easily check that $L$ can be represented as
      $$L = a(r-1)+br$$
      where $a, b in mathbb{Z}$ are positive integers and
      $$K = a + b$$
      is even."



      My first attempt was of course to apply something like the euclidean algorithm, which gives us
      $$L = k(r-1)+m$$
      where $k, m$ are positive integers, $m<r-1$ (I tried to apply euclidean algorithm for $r$ instead of $r-1$ at first, but this seems to be more useful...)



      As we have
      $$L = k(r-1) + m geq 2(r-1)²$$
      with $m leq r-2$, we know
      $$k+1-m-2 geq 2r(r-1) -(r-2)-2 geq 2r² -3r -1 geq (r-1)²$$
      for all $rgeq 2$. (The case $r=1$ is not interesting.)



      Therefore we can write
      $$L = (k+1)(r-1)+m-r+1 = (k+1-m-2)(r-1)+(m+1)r$$
      where now $(k+1-m-2), (m+1)$ are positive integers. But their sum equals
      $$(k+1-m-2)+(m+1)=k$$
      which of course does not have to be even.



      Can somebody give me a hint how to manipulate this way of writing
      $$L=a(r-1)+br$$
      where $a+b$ is even?



      This whole thing is a little weird, if you consider e.g. $r=5, L=33geq 32$, you could write
      $$33=5r+2(r-1)$$
      where $5+2$ is not even, but you also can write
      $$33=r+7(r-1)$$
      where $1+7$ is even (I don't know if an expert in stuff like this would consider this as weird, but I do...)



      Thanks for any help :)






      summation integers euclidean-algorithm







      share|cite|improve this question









      $endgroup$




      Let $r in mathbb{N}$ be a natural number. Let
      $$L geq 2(r-1)²$$
      A paper (on quantum information theory, I'm not an expert in number theory or so...) I'm recently reading now says



      "One can easily check that $L$ can be represented as
      $$L = a(r-1)+br$$
      where $a, b in mathbb{Z}$ are positive integers and
      $$K = a + b$$
      is even."



      My first attempt was of course to apply something like the euclidean algorithm, which gives us
      $$L = k(r-1)+m$$
      where $k, m$ are positive integers, $m<r-1$ (I tried to apply euclidean algorithm for $r$ instead of $r-1$ at first, but this seems to be more useful...)



      As we have
      $$L = k(r-1) + m geq 2(r-1)²$$
      with $m leq r-2$, we know
      $$k+1-m-2 geq 2r(r-1) -(r-2)-2 geq 2r² -3r -1 geq (r-1)²$$
      for all $rgeq 2$. (The case $r=1$ is not interesting.)



      Therefore we can write
      $$L = (k+1)(r-1)+m-r+1 = (k+1-m-2)(r-1)+(m+1)r$$
      where now $(k+1-m-2), (m+1)$ are positive integers. But their sum equals
      $$(k+1-m-2)+(m+1)=k$$
      which of course does not have to be even.



      Can somebody give me a hint how to manipulate this way of writing
      $$L=a(r-1)+br$$
      where $a+b$ is even?



      This whole thing is a little weird, if you consider e.g. $r=5, L=33geq 32$, you could write
      $$33=5r+2(r-1)$$
      where $5+2$ is not even, but you also can write
      $$33=r+7(r-1)$$
      where $1+7$ is even (I don't know if an expert in stuff like this would consider this as weird, but I do...)



      Thanks for any help :)






      summation integers euclidean-algorithm




      summation integers euclidean-algorithm






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 22 '18 at 23:29









      Fritz HefterFritz Hefter

      264




      264






















          1 Answer
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          $begingroup$

          While I'm not sure how to make the OP's idea work, the following is a solution to the problem he poses.



          First suppose $(r-1) nmid L$. Notice that the condition that $a,b$ exist so that $L = a(r-1) + br$ is equivalent to saying that $a,b$ exist so that $r-1 = frac{L-b}{a + b}$. Choose $b$ so that $L-b = 2k(r-1)$, where $k$ is chosen so that $2k(r-1)$ is the largest multiple of $2(r-1)$ less than or equal to $L$. Then we must have $a + b = 2k$ and that $0 < b leq r-1$. Because $k geq r-1$, it follows that $a > 0$.



          Now suppose $L = m(r-1)$ for some odd number $m$. Then take $b = r-1$ and $a = m-r$.



          Now suppose $L = 2m(r-1)$ for some odd number $m$. Then take $b = 2(r-1)$ and $a = 2(m-r)$.



          That should be all the cases.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            0












            $begingroup$

            While I'm not sure how to make the OP's idea work, the following is a solution to the problem he poses.



            First suppose $(r-1) nmid L$. Notice that the condition that $a,b$ exist so that $L = a(r-1) + br$ is equivalent to saying that $a,b$ exist so that $r-1 = frac{L-b}{a + b}$. Choose $b$ so that $L-b = 2k(r-1)$, where $k$ is chosen so that $2k(r-1)$ is the largest multiple of $2(r-1)$ less than or equal to $L$. Then we must have $a + b = 2k$ and that $0 < b leq r-1$. Because $k geq r-1$, it follows that $a > 0$.



            Now suppose $L = m(r-1)$ for some odd number $m$. Then take $b = r-1$ and $a = m-r$.



            Now suppose $L = 2m(r-1)$ for some odd number $m$. Then take $b = 2(r-1)$ and $a = 2(m-r)$.



            That should be all the cases.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              While I'm not sure how to make the OP's idea work, the following is a solution to the problem he poses.



              First suppose $(r-1) nmid L$. Notice that the condition that $a,b$ exist so that $L = a(r-1) + br$ is equivalent to saying that $a,b$ exist so that $r-1 = frac{L-b}{a + b}$. Choose $b$ so that $L-b = 2k(r-1)$, where $k$ is chosen so that $2k(r-1)$ is the largest multiple of $2(r-1)$ less than or equal to $L$. Then we must have $a + b = 2k$ and that $0 < b leq r-1$. Because $k geq r-1$, it follows that $a > 0$.



              Now suppose $L = m(r-1)$ for some odd number $m$. Then take $b = r-1$ and $a = m-r$.



              Now suppose $L = 2m(r-1)$ for some odd number $m$. Then take $b = 2(r-1)$ and $a = 2(m-r)$.



              That should be all the cases.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                While I'm not sure how to make the OP's idea work, the following is a solution to the problem he poses.



                First suppose $(r-1) nmid L$. Notice that the condition that $a,b$ exist so that $L = a(r-1) + br$ is equivalent to saying that $a,b$ exist so that $r-1 = frac{L-b}{a + b}$. Choose $b$ so that $L-b = 2k(r-1)$, where $k$ is chosen so that $2k(r-1)$ is the largest multiple of $2(r-1)$ less than or equal to $L$. Then we must have $a + b = 2k$ and that $0 < b leq r-1$. Because $k geq r-1$, it follows that $a > 0$.



                Now suppose $L = m(r-1)$ for some odd number $m$. Then take $b = r-1$ and $a = m-r$.



                Now suppose $L = 2m(r-1)$ for some odd number $m$. Then take $b = 2(r-1)$ and $a = 2(m-r)$.



                That should be all the cases.






                share|cite|improve this answer











                $endgroup$



                While I'm not sure how to make the OP's idea work, the following is a solution to the problem he poses.



                First suppose $(r-1) nmid L$. Notice that the condition that $a,b$ exist so that $L = a(r-1) + br$ is equivalent to saying that $a,b$ exist so that $r-1 = frac{L-b}{a + b}$. Choose $b$ so that $L-b = 2k(r-1)$, where $k$ is chosen so that $2k(r-1)$ is the largest multiple of $2(r-1)$ less than or equal to $L$. Then we must have $a + b = 2k$ and that $0 < b leq r-1$. Because $k geq r-1$, it follows that $a > 0$.



                Now suppose $L = m(r-1)$ for some odd number $m$. Then take $b = r-1$ and $a = m-r$.



                Now suppose $L = 2m(r-1)$ for some odd number $m$. Then take $b = 2(r-1)$ and $a = 2(m-r)$.



                That should be all the cases.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 23 '18 at 1:29

























                answered Dec 23 '18 at 0:10









                Ashvin SwaminathanAshvin Swaminathan

                1,571520




                1,571520






























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