Determine all $a$ and $b$ natural numbers such that $frac {a^2+2b} {b^2-2a}$ and $frac {b^2+2a} {a^2-2b}$ are...












2












$begingroup$


I proceeded in the following way:



It is clear that $a ne 0$ and $b ne 0$.



Let $frac {a^2+2b} {b^2-2a} = k, k in mathbb{Z} tag 1$
and $frac {b^2+2a} {a^2-2b} = m, m in mathbb{Z} tag 2$



1. If $a = b$:

Let $a = b = n$. Then $k = m = frac {n^2+2n} {n^2-2n} = frac {n(n+2)} {n(n-2)} = frac {n+2} {n-2} = p, p in mathbb{Z}$. From here we get that $n in {1, 3, 4, 6}$. I don't know if these are all so I can't prove that there are no others.



2. If $a ne b$:

For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.

From $(1)$ we get that $a^2+2b ge b^2-2a implies a+2 ge b tag {3}$

From $(2)$ we get that $b^2+2a ge a^2-2b implies b+2 ge a tag {4}$
Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.






fractions natural-numbers







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I proceeded in the following way:



    It is clear that $a ne 0$ and $b ne 0$.



    Let $frac {a^2+2b} {b^2-2a} = k, k in mathbb{Z} tag 1$
    and $frac {b^2+2a} {a^2-2b} = m, m in mathbb{Z} tag 2$



    1. If $a = b$:

    Let $a = b = n$. Then $k = m = frac {n^2+2n} {n^2-2n} = frac {n(n+2)} {n(n-2)} = frac {n+2} {n-2} = p, p in mathbb{Z}$. From here we get that $n in {1, 3, 4, 6}$. I don't know if these are all so I can't prove that there are no others.



    2. If $a ne b$:

    For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.

    From $(1)$ we get that $a^2+2b ge b^2-2a implies a+2 ge b tag {3}$

    From $(2)$ we get that $b^2+2a ge a^2-2b implies b+2 ge a tag {4}$
    Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.






    fractions natural-numbers







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I proceeded in the following way:



      It is clear that $a ne 0$ and $b ne 0$.



      Let $frac {a^2+2b} {b^2-2a} = k, k in mathbb{Z} tag 1$
      and $frac {b^2+2a} {a^2-2b} = m, m in mathbb{Z} tag 2$



      1. If $a = b$:

      Let $a = b = n$. Then $k = m = frac {n^2+2n} {n^2-2n} = frac {n(n+2)} {n(n-2)} = frac {n+2} {n-2} = p, p in mathbb{Z}$. From here we get that $n in {1, 3, 4, 6}$. I don't know if these are all so I can't prove that there are no others.



      2. If $a ne b$:

      For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.

      From $(1)$ we get that $a^2+2b ge b^2-2a implies a+2 ge b tag {3}$

      From $(2)$ we get that $b^2+2a ge a^2-2b implies b+2 ge a tag {4}$
      Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.






      fractions natural-numbers







      share|cite|improve this question











      $endgroup$




      I proceeded in the following way:



      It is clear that $a ne 0$ and $b ne 0$.



      Let $frac {a^2+2b} {b^2-2a} = k, k in mathbb{Z} tag 1$
      and $frac {b^2+2a} {a^2-2b} = m, m in mathbb{Z} tag 2$



      1. If $a = b$:

      Let $a = b = n$. Then $k = m = frac {n^2+2n} {n^2-2n} = frac {n(n+2)} {n(n-2)} = frac {n+2} {n-2} = p, p in mathbb{Z}$. From here we get that $n in {1, 3, 4, 6}$. I don't know if these are all so I can't prove that there are no others.



      2. If $a ne b$:

      For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.

      From $(1)$ we get that $a^2+2b ge b^2-2a implies a+2 ge b tag {3}$

      From $(2)$ we get that $b^2+2a ge a^2-2b implies b+2 ge a tag {4}$
      Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.






      fractions natural-numbers




      fractions natural-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 19:26







      Krisztián Kiss

















      asked Jan 15 at 19:18









      Krisztián KissKrisztián Kiss

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      484






















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          $begingroup$

          So we have $bin {a-2,a-1,a,a+1,a+2}$



          Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
          so $$a^2-6a+4mid a^2+2a-4$$



          and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$



          so $$ (a-7)^2leq 37implies |a-7|leq 6...$$



          We do similary for all other cases...






          share|cite|improve this answer









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            $begingroup$

            So we have $bin {a-2,a-1,a,a+1,a+2}$



            Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
            so $$a^2-6a+4mid a^2+2a-4$$



            and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$



            so $$ (a-7)^2leq 37implies |a-7|leq 6...$$



            We do similary for all other cases...






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              So we have $bin {a-2,a-1,a,a+1,a+2}$



              Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
              so $$a^2-6a+4mid a^2+2a-4$$



              and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$



              so $$ (a-7)^2leq 37implies |a-7|leq 6...$$



              We do similary for all other cases...






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                So we have $bin {a-2,a-1,a,a+1,a+2}$



                Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
                so $$a^2-6a+4mid a^2+2a-4$$



                and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$



                so $$ (a-7)^2leq 37implies |a-7|leq 6...$$



                We do similary for all other cases...






                share|cite|improve this answer









                $endgroup$



                So we have $bin {a-2,a-1,a,a+1,a+2}$



                Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
                so $$a^2-6a+4mid a^2+2a-4$$



                and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$



                so $$ (a-7)^2leq 37implies |a-7|leq 6...$$



                We do similary for all other cases...







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 19:37









                greedoidgreedoid

                43.2k1153105




                43.2k1153105






























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