Prove $x^n+y^n=z^n$ has no solution in natural numbers for $ngeq z$












1












$begingroup$


If $ x$, $ y$, $ z$, and $ n$ are natural numbers, and $ ngeq z$ then prove that the relation $ x^n + y^n = z^n$ does not hold.



I have a question: Can I use Fermats Last Theorem here directly? It makes my life easier, but I have a doubt whether or not in Olympiads (if at all they ask such a question; most probably they would not :D) I can directly state :By Fermats Last Theorem... and then proceed, for till now I dont think I have come across a simple and elegant proof for the theorem. I would like to have suggestions on this. Please help. Any other solution is also welcome.






number-theory







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$endgroup$












  • $begingroup$
    I think it goes against the purpose and spirit of the problem to use FLT. So my guess is no, you can't use it.
    $endgroup$
    – Arthur
    Jan 8 at 19:06












  • $begingroup$
    Fermat's Last Theorem may come in use here, but what happens if $n = 1$, or $n=2$? (due to the fact that $nge z$).
    $endgroup$
    – Decaf-Math
    Jan 8 at 19:06
















1












$begingroup$


If $ x$, $ y$, $ z$, and $ n$ are natural numbers, and $ ngeq z$ then prove that the relation $ x^n + y^n = z^n$ does not hold.



I have a question: Can I use Fermats Last Theorem here directly? It makes my life easier, but I have a doubt whether or not in Olympiads (if at all they ask such a question; most probably they would not :D) I can directly state :By Fermats Last Theorem... and then proceed, for till now I dont think I have come across a simple and elegant proof for the theorem. I would like to have suggestions on this. Please help. Any other solution is also welcome.






number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think it goes against the purpose and spirit of the problem to use FLT. So my guess is no, you can't use it.
    $endgroup$
    – Arthur
    Jan 8 at 19:06












  • $begingroup$
    Fermat's Last Theorem may come in use here, but what happens if $n = 1$, or $n=2$? (due to the fact that $nge z$).
    $endgroup$
    – Decaf-Math
    Jan 8 at 19:06














1












1








1





$begingroup$


If $ x$, $ y$, $ z$, and $ n$ are natural numbers, and $ ngeq z$ then prove that the relation $ x^n + y^n = z^n$ does not hold.



I have a question: Can I use Fermats Last Theorem here directly? It makes my life easier, but I have a doubt whether or not in Olympiads (if at all they ask such a question; most probably they would not :D) I can directly state :By Fermats Last Theorem... and then proceed, for till now I dont think I have come across a simple and elegant proof for the theorem. I would like to have suggestions on this. Please help. Any other solution is also welcome.






number-theory







share|cite|improve this question









$endgroup$




If $ x$, $ y$, $ z$, and $ n$ are natural numbers, and $ ngeq z$ then prove that the relation $ x^n + y^n = z^n$ does not hold.



I have a question: Can I use Fermats Last Theorem here directly? It makes my life easier, but I have a doubt whether or not in Olympiads (if at all they ask such a question; most probably they would not :D) I can directly state :By Fermats Last Theorem... and then proceed, for till now I dont think I have come across a simple and elegant proof for the theorem. I would like to have suggestions on this. Please help. Any other solution is also welcome.






number-theory




number-theory






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 19:00









YellowYellow

16011




16011












  • $begingroup$
    I think it goes against the purpose and spirit of the problem to use FLT. So my guess is no, you can't use it.
    $endgroup$
    – Arthur
    Jan 8 at 19:06












  • $begingroup$
    Fermat's Last Theorem may come in use here, but what happens if $n = 1$, or $n=2$? (due to the fact that $nge z$).
    $endgroup$
    – Decaf-Math
    Jan 8 at 19:06


















  • $begingroup$
    I think it goes against the purpose and spirit of the problem to use FLT. So my guess is no, you can't use it.
    $endgroup$
    – Arthur
    Jan 8 at 19:06












  • $begingroup$
    Fermat's Last Theorem may come in use here, but what happens if $n = 1$, or $n=2$? (due to the fact that $nge z$).
    $endgroup$
    – Decaf-Math
    Jan 8 at 19:06
















$begingroup$
I think it goes against the purpose and spirit of the problem to use FLT. So my guess is no, you can't use it.
$endgroup$
– Arthur
Jan 8 at 19:06






$begingroup$
I think it goes against the purpose and spirit of the problem to use FLT. So my guess is no, you can't use it.
$endgroup$
– Arthur
Jan 8 at 19:06














$begingroup$
Fermat's Last Theorem may come in use here, but what happens if $n = 1$, or $n=2$? (due to the fact that $nge z$).
$endgroup$
– Decaf-Math
Jan 8 at 19:06




$begingroup$
Fermat's Last Theorem may come in use here, but what happens if $n = 1$, or $n=2$? (due to the fact that $nge z$).
$endgroup$
– Decaf-Math
Jan 8 at 19:06










3 Answers
3






active

oldest

votes


















1












$begingroup$

I think the object of the question is to prove that $x,y,z,nin mathbb N wedge x^n+y^n=z^nRightarrow zge 3$, and hence $nge 3$. Having proven this much, invoking FLT would be appropriate.



Also, the question must be using the (general but not universal) convention $0not in mathbb N$ since $0^n+0^n=0^n$ for all $nge 1$, making $n>z=0$.



So if $x,y ge 1$ then $x^n+y^nge 2$ requiring $zge 2$. Although $1^1+1^1=2^1$, this does not satisfy $nge z$. Also, $1^2+1^2ne 2^2$. If $x>1$ or $y>1$, then $z>2 Rightarrow nge 3$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Here’s a proof without using FLT. Without loss of generality assume $x leq y$. Clearly $z>y$, so $n geq z geq y+1$. Then $$z^n geq (1+y)^n geq y^n+ny^{n-1} geq y^n+(1+y)y^{n-1}>2y^n geq x^n+y^n$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You say $z^n geq x^n + y^n$. Then?
      $endgroup$
      – Yellow
      Jan 9 at 3:42



















    3












    $begingroup$

    Mathematically, using Fermat's Last Theorem would be a valid proof. Of course, you need to consider the case $z=1$ and $z=2$ separately, because FLT needs $n > 2$, and you have $n > z$. If they were to ever give this problem in IMO, an FLT solution should receive full points (and that's why they'd never give such a problem there).



    Of course, if this is a practice problem, and your goal is to learn number theory rather than produce a proof, you probably should only use theorems that you can prove.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was always told that IMO rules granted no points for the use of theorems (especially very advanced/recent ones) that did the whole job for you.
      $endgroup$
      – Mindlack
      Jan 8 at 19:42










    • $begingroup$
      Yup, same here @Mindlack.
      $endgroup$
      – Yellow
      Jan 9 at 3:37











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    3 Answers
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    3 Answers
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    1












    $begingroup$

    I think the object of the question is to prove that $x,y,z,nin mathbb N wedge x^n+y^n=z^nRightarrow zge 3$, and hence $nge 3$. Having proven this much, invoking FLT would be appropriate.



    Also, the question must be using the (general but not universal) convention $0not in mathbb N$ since $0^n+0^n=0^n$ for all $nge 1$, making $n>z=0$.



    So if $x,y ge 1$ then $x^n+y^nge 2$ requiring $zge 2$. Although $1^1+1^1=2^1$, this does not satisfy $nge z$. Also, $1^2+1^2ne 2^2$. If $x>1$ or $y>1$, then $z>2 Rightarrow nge 3$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I think the object of the question is to prove that $x,y,z,nin mathbb N wedge x^n+y^n=z^nRightarrow zge 3$, and hence $nge 3$. Having proven this much, invoking FLT would be appropriate.



      Also, the question must be using the (general but not universal) convention $0not in mathbb N$ since $0^n+0^n=0^n$ for all $nge 1$, making $n>z=0$.



      So if $x,y ge 1$ then $x^n+y^nge 2$ requiring $zge 2$. Although $1^1+1^1=2^1$, this does not satisfy $nge z$. Also, $1^2+1^2ne 2^2$. If $x>1$ or $y>1$, then $z>2 Rightarrow nge 3$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I think the object of the question is to prove that $x,y,z,nin mathbb N wedge x^n+y^n=z^nRightarrow zge 3$, and hence $nge 3$. Having proven this much, invoking FLT would be appropriate.



        Also, the question must be using the (general but not universal) convention $0not in mathbb N$ since $0^n+0^n=0^n$ for all $nge 1$, making $n>z=0$.



        So if $x,y ge 1$ then $x^n+y^nge 2$ requiring $zge 2$. Although $1^1+1^1=2^1$, this does not satisfy $nge z$. Also, $1^2+1^2ne 2^2$. If $x>1$ or $y>1$, then $z>2 Rightarrow nge 3$.






        share|cite|improve this answer









        $endgroup$



        I think the object of the question is to prove that $x,y,z,nin mathbb N wedge x^n+y^n=z^nRightarrow zge 3$, and hence $nge 3$. Having proven this much, invoking FLT would be appropriate.



        Also, the question must be using the (general but not universal) convention $0not in mathbb N$ since $0^n+0^n=0^n$ for all $nge 1$, making $n>z=0$.



        So if $x,y ge 1$ then $x^n+y^nge 2$ requiring $zge 2$. Although $1^1+1^1=2^1$, this does not satisfy $nge z$. Also, $1^2+1^2ne 2^2$. If $x>1$ or $y>1$, then $z>2 Rightarrow nge 3$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 21:00









        Keith BackmanKeith Backman

        1,1141712




        1,1141712























            4












            $begingroup$

            Here’s a proof without using FLT. Without loss of generality assume $x leq y$. Clearly $z>y$, so $n geq z geq y+1$. Then $$z^n geq (1+y)^n geq y^n+ny^{n-1} geq y^n+(1+y)y^{n-1}>2y^n geq x^n+y^n$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You say $z^n geq x^n + y^n$. Then?
              $endgroup$
              – Yellow
              Jan 9 at 3:42
















            4












            $begingroup$

            Here’s a proof without using FLT. Without loss of generality assume $x leq y$. Clearly $z>y$, so $n geq z geq y+1$. Then $$z^n geq (1+y)^n geq y^n+ny^{n-1} geq y^n+(1+y)y^{n-1}>2y^n geq x^n+y^n$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You say $z^n geq x^n + y^n$. Then?
              $endgroup$
              – Yellow
              Jan 9 at 3:42














            4












            4








            4





            $begingroup$

            Here’s a proof without using FLT. Without loss of generality assume $x leq y$. Clearly $z>y$, so $n geq z geq y+1$. Then $$z^n geq (1+y)^n geq y^n+ny^{n-1} geq y^n+(1+y)y^{n-1}>2y^n geq x^n+y^n$$






            share|cite|improve this answer









            $endgroup$



            Here’s a proof without using FLT. Without loss of generality assume $x leq y$. Clearly $z>y$, so $n geq z geq y+1$. Then $$z^n geq (1+y)^n geq y^n+ny^{n-1} geq y^n+(1+y)y^{n-1}>2y^n geq x^n+y^n$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 19:51









            user632896user632896

            411




            411












            • $begingroup$
              You say $z^n geq x^n + y^n$. Then?
              $endgroup$
              – Yellow
              Jan 9 at 3:42


















            • $begingroup$
              You say $z^n geq x^n + y^n$. Then?
              $endgroup$
              – Yellow
              Jan 9 at 3:42
















            $begingroup$
            You say $z^n geq x^n + y^n$. Then?
            $endgroup$
            – Yellow
            Jan 9 at 3:42




            $begingroup$
            You say $z^n geq x^n + y^n$. Then?
            $endgroup$
            – Yellow
            Jan 9 at 3:42











            3












            $begingroup$

            Mathematically, using Fermat's Last Theorem would be a valid proof. Of course, you need to consider the case $z=1$ and $z=2$ separately, because FLT needs $n > 2$, and you have $n > z$. If they were to ever give this problem in IMO, an FLT solution should receive full points (and that's why they'd never give such a problem there).



            Of course, if this is a practice problem, and your goal is to learn number theory rather than produce a proof, you probably should only use theorems that you can prove.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I was always told that IMO rules granted no points for the use of theorems (especially very advanced/recent ones) that did the whole job for you.
              $endgroup$
              – Mindlack
              Jan 8 at 19:42










            • $begingroup$
              Yup, same here @Mindlack.
              $endgroup$
              – Yellow
              Jan 9 at 3:37
















            3












            $begingroup$

            Mathematically, using Fermat's Last Theorem would be a valid proof. Of course, you need to consider the case $z=1$ and $z=2$ separately, because FLT needs $n > 2$, and you have $n > z$. If they were to ever give this problem in IMO, an FLT solution should receive full points (and that's why they'd never give such a problem there).



            Of course, if this is a practice problem, and your goal is to learn number theory rather than produce a proof, you probably should only use theorems that you can prove.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I was always told that IMO rules granted no points for the use of theorems (especially very advanced/recent ones) that did the whole job for you.
              $endgroup$
              – Mindlack
              Jan 8 at 19:42










            • $begingroup$
              Yup, same here @Mindlack.
              $endgroup$
              – Yellow
              Jan 9 at 3:37














            3












            3








            3





            $begingroup$

            Mathematically, using Fermat's Last Theorem would be a valid proof. Of course, you need to consider the case $z=1$ and $z=2$ separately, because FLT needs $n > 2$, and you have $n > z$. If they were to ever give this problem in IMO, an FLT solution should receive full points (and that's why they'd never give such a problem there).



            Of course, if this is a practice problem, and your goal is to learn number theory rather than produce a proof, you probably should only use theorems that you can prove.






            share|cite|improve this answer









            $endgroup$



            Mathematically, using Fermat's Last Theorem would be a valid proof. Of course, you need to consider the case $z=1$ and $z=2$ separately, because FLT needs $n > 2$, and you have $n > z$. If they were to ever give this problem in IMO, an FLT solution should receive full points (and that's why they'd never give such a problem there).



            Of course, if this is a practice problem, and your goal is to learn number theory rather than produce a proof, you probably should only use theorems that you can prove.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 19:15









            Todor MarkovTodor Markov

            1,819410




            1,819410












            • $begingroup$
              I was always told that IMO rules granted no points for the use of theorems (especially very advanced/recent ones) that did the whole job for you.
              $endgroup$
              – Mindlack
              Jan 8 at 19:42










            • $begingroup$
              Yup, same here @Mindlack.
              $endgroup$
              – Yellow
              Jan 9 at 3:37


















            • $begingroup$
              I was always told that IMO rules granted no points for the use of theorems (especially very advanced/recent ones) that did the whole job for you.
              $endgroup$
              – Mindlack
              Jan 8 at 19:42










            • $begingroup$
              Yup, same here @Mindlack.
              $endgroup$
              – Yellow
              Jan 9 at 3:37
















            $begingroup$
            I was always told that IMO rules granted no points for the use of theorems (especially very advanced/recent ones) that did the whole job for you.
            $endgroup$
            – Mindlack
            Jan 8 at 19:42




            $begingroup$
            I was always told that IMO rules granted no points for the use of theorems (especially very advanced/recent ones) that did the whole job for you.
            $endgroup$
            – Mindlack
            Jan 8 at 19:42












            $begingroup$
            Yup, same here @Mindlack.
            $endgroup$
            – Yellow
            Jan 9 at 3:37




            $begingroup$
            Yup, same here @Mindlack.
            $endgroup$
            – Yellow
            Jan 9 at 3:37


















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