An analogue to Euler's sum of powers conjecture












4












$begingroup$


What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \
vdots $$
we instead considered
$$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \
a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \
vdots $$ My gut feeling is that this will have infinitely many solutions.



Note: $a_i,b_i$ are all positive integers.






elementary-number-theory







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$endgroup$

















    4












    $begingroup$


    What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \
    a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \
    vdots $$
    we instead considered
    $$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \
    a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \
    a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \
    vdots $$     My gut feeling is that this will have infinitely many solutions.



    Note: $a_i,b_i$ are all positive integers.






    elementary-number-theory







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      0



      $begingroup$


      What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \
      a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \
      vdots $$
      we instead considered
      $$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \
      a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \
      a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \
      vdots $$       My gut feeling is that this will have infinitely many solutions.



      Note: $a_i,b_i$ are all positive integers.






      elementary-number-theory







      share|cite|improve this question











      $endgroup$




      What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \
      a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \
      vdots $$
      we instead considered
      $$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \
      a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \
      a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \
      vdots $$       My gut feeling is that this will have infinitely many solutions.



      Note: $a_i,b_i$ are all positive integers.






      elementary-number-theory




      elementary-number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 11 at 7:23







      Stupid Questions Inc

















      asked Jan 10 at 11:24









      Stupid Questions IncStupid Questions Inc

      7010




      7010






















          1 Answer
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          2












          $begingroup$

          Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
          $$
          (z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
          $$
          for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
          If you only want positive solutions, it is also possible because we have
          $$
          1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
          $$
          Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
          $$
          x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
          $$
          has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
          $$
          (y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
          $$
          where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
            $endgroup$
            – Adam Bailey
            Jan 10 at 22:43










          • $begingroup$
            Does your strategy work if we restrict everything to the naturals?
            $endgroup$
            – Stupid Questions Inc
            Jan 13 at 19:50










          • $begingroup$
            @StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
            $endgroup$
            – Seewoo Lee
            Jan 13 at 22:06











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          2












          $begingroup$

          Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
          $$
          (z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
          $$
          for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
          If you only want positive solutions, it is also possible because we have
          $$
          1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
          $$
          Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
          $$
          x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
          $$
          has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
          $$
          (y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
          $$
          where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
            $endgroup$
            – Adam Bailey
            Jan 10 at 22:43










          • $begingroup$
            Does your strategy work if we restrict everything to the naturals?
            $endgroup$
            – Stupid Questions Inc
            Jan 13 at 19:50










          • $begingroup$
            @StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
            $endgroup$
            – Seewoo Lee
            Jan 13 at 22:06
















          2












          $begingroup$

          Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
          $$
          (z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
          $$
          for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
          If you only want positive solutions, it is also possible because we have
          $$
          1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
          $$
          Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
          $$
          x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
          $$
          has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
          $$
          (y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
          $$
          where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
            $endgroup$
            – Adam Bailey
            Jan 10 at 22:43










          • $begingroup$
            Does your strategy work if we restrict everything to the naturals?
            $endgroup$
            – Stupid Questions Inc
            Jan 13 at 19:50










          • $begingroup$
            @StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
            $endgroup$
            – Seewoo Lee
            Jan 13 at 22:06














          2












          2








          2





          $begingroup$

          Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
          $$
          (z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
          $$
          for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
          If you only want positive solutions, it is also possible because we have
          $$
          1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
          $$
          Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
          $$
          x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
          $$
          has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
          $$
          (y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
          $$
          where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.






          share|cite|improve this answer









          $endgroup$



          Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
          $$
          (z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
          $$
          for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
          If you only want positive solutions, it is also possible because we have
          $$
          1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
          $$
          Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
          $$
          x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
          $$
          has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
          $$
          (y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
          $$
          where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 13:11









          Seewoo LeeSeewoo Lee

          6,616926




          6,616926












          • $begingroup$
            +1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
            $endgroup$
            – Adam Bailey
            Jan 10 at 22:43










          • $begingroup$
            Does your strategy work if we restrict everything to the naturals?
            $endgroup$
            – Stupid Questions Inc
            Jan 13 at 19:50










          • $begingroup$
            @StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
            $endgroup$
            – Seewoo Lee
            Jan 13 at 22:06


















          • $begingroup$
            +1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
            $endgroup$
            – Adam Bailey
            Jan 10 at 22:43










          • $begingroup$
            Does your strategy work if we restrict everything to the naturals?
            $endgroup$
            – Stupid Questions Inc
            Jan 13 at 19:50










          • $begingroup$
            @StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
            $endgroup$
            – Seewoo Lee
            Jan 13 at 22:06
















          $begingroup$
          +1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
          $endgroup$
          – Adam Bailey
          Jan 10 at 22:43




          $begingroup$
          +1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
          $endgroup$
          – Adam Bailey
          Jan 10 at 22:43












          $begingroup$
          Does your strategy work if we restrict everything to the naturals?
          $endgroup$
          – Stupid Questions Inc
          Jan 13 at 19:50




          $begingroup$
          Does your strategy work if we restrict everything to the naturals?
          $endgroup$
          – Stupid Questions Inc
          Jan 13 at 19:50












          $begingroup$
          @StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
          $endgroup$
          – Seewoo Lee
          Jan 13 at 22:06




          $begingroup$
          @StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
          $endgroup$
          – Seewoo Lee
          Jan 13 at 22:06


















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