Is there a gap in this proof of “$a^3+b^3not=c^3$”











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My friend pass me a simple proof that there is no positive integer solution for equation $a^3+b^3=c^3$.



I'm not sure whether the proof is right or not.



The proof:



suppose $a,b,c$ are positive integers , coprime pairwisely and satisfy the equation $a^3+b^3=c^3$



then we get



$(a+b-c)^3=3(a+b)(c-a)(c-b)~~~~~~(1)$



let $x=a+b$, so $x$ divides $c^3$ ,since $c^3 = a^3+b^3=(a+b)(a^2+ab+c^2)$



so, let $y=c-b,z=c-a$,then $y$ divides $a^3$, then $z$ divides $b^3$



we have $x,y,z$ are pairwisely coprime since $a^3,b^3,c^3$ are pairwisely coprime



by $(1)$, we can see



$(x-y-z)^3=8(a+b-c)^3=24xyz~~~~~(2)$



then $24xyz$ should have factor $x^3$, so $x^2$ divides $24$



Then $x=1$ or $2$



It's imposible.










share|cite|improve this question
























  • @ Dietrich Burde ,but in the begining we assume $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:21










  • $(1)$ should be $(a+b-c)^3-3(a+b)(c-a)(c-b)=c^3-a^3-b^3$.
    – Dietrich Burde
    Nov 18 at 9:21












  • @DietrichBurde (1) is true assuming that $a^3+b^3=c^3$?
    – Oldboy
    Nov 18 at 9:21






  • 1




    @DietrichBurde Similar question is OK, but Letingok asks what's wrong with his proof which seems to be 10 times shorter.
    – Oldboy
    Nov 18 at 9:30






  • 3




    why should $x^3$ divide $24xyz$ ?
    – mercio
    Nov 18 at 9:35















up vote
0
down vote

favorite












My friend pass me a simple proof that there is no positive integer solution for equation $a^3+b^3=c^3$.



I'm not sure whether the proof is right or not.



The proof:



suppose $a,b,c$ are positive integers , coprime pairwisely and satisfy the equation $a^3+b^3=c^3$



then we get



$(a+b-c)^3=3(a+b)(c-a)(c-b)~~~~~~(1)$



let $x=a+b$, so $x$ divides $c^3$ ,since $c^3 = a^3+b^3=(a+b)(a^2+ab+c^2)$



so, let $y=c-b,z=c-a$,then $y$ divides $a^3$, then $z$ divides $b^3$



we have $x,y,z$ are pairwisely coprime since $a^3,b^3,c^3$ are pairwisely coprime



by $(1)$, we can see



$(x-y-z)^3=8(a+b-c)^3=24xyz~~~~~(2)$



then $24xyz$ should have factor $x^3$, so $x^2$ divides $24$



Then $x=1$ or $2$



It's imposible.










share|cite|improve this question
























  • @ Dietrich Burde ,but in the begining we assume $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:21










  • $(1)$ should be $(a+b-c)^3-3(a+b)(c-a)(c-b)=c^3-a^3-b^3$.
    – Dietrich Burde
    Nov 18 at 9:21












  • @DietrichBurde (1) is true assuming that $a^3+b^3=c^3$?
    – Oldboy
    Nov 18 at 9:21






  • 1




    @DietrichBurde Similar question is OK, but Letingok asks what's wrong with his proof which seems to be 10 times shorter.
    – Oldboy
    Nov 18 at 9:30






  • 3




    why should $x^3$ divide $24xyz$ ?
    – mercio
    Nov 18 at 9:35













up vote
0
down vote

favorite









up vote
0
down vote

favorite











My friend pass me a simple proof that there is no positive integer solution for equation $a^3+b^3=c^3$.



I'm not sure whether the proof is right or not.



The proof:



suppose $a,b,c$ are positive integers , coprime pairwisely and satisfy the equation $a^3+b^3=c^3$



then we get



$(a+b-c)^3=3(a+b)(c-a)(c-b)~~~~~~(1)$



let $x=a+b$, so $x$ divides $c^3$ ,since $c^3 = a^3+b^3=(a+b)(a^2+ab+c^2)$



so, let $y=c-b,z=c-a$,then $y$ divides $a^3$, then $z$ divides $b^3$



we have $x,y,z$ are pairwisely coprime since $a^3,b^3,c^3$ are pairwisely coprime



by $(1)$, we can see



$(x-y-z)^3=8(a+b-c)^3=24xyz~~~~~(2)$



then $24xyz$ should have factor $x^3$, so $x^2$ divides $24$



Then $x=1$ or $2$



It's imposible.










share|cite|improve this question















My friend pass me a simple proof that there is no positive integer solution for equation $a^3+b^3=c^3$.



I'm not sure whether the proof is right or not.



The proof:



suppose $a,b,c$ are positive integers , coprime pairwisely and satisfy the equation $a^3+b^3=c^3$



then we get



$(a+b-c)^3=3(a+b)(c-a)(c-b)~~~~~~(1)$



let $x=a+b$, so $x$ divides $c^3$ ,since $c^3 = a^3+b^3=(a+b)(a^2+ab+c^2)$



so, let $y=c-b,z=c-a$,then $y$ divides $a^3$, then $z$ divides $b^3$



we have $x,y,z$ are pairwisely coprime since $a^3,b^3,c^3$ are pairwisely coprime



by $(1)$, we can see



$(x-y-z)^3=8(a+b-c)^3=24xyz~~~~~(2)$



then $24xyz$ should have factor $x^3$, so $x^2$ divides $24$



Then $x=1$ or $2$



It's imposible.







elementary-number-theory






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edited Nov 18 at 9:15

























asked Nov 18 at 9:06









Leitingok

941932




941932












  • @ Dietrich Burde ,but in the begining we assume $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:21










  • $(1)$ should be $(a+b-c)^3-3(a+b)(c-a)(c-b)=c^3-a^3-b^3$.
    – Dietrich Burde
    Nov 18 at 9:21












  • @DietrichBurde (1) is true assuming that $a^3+b^3=c^3$?
    – Oldboy
    Nov 18 at 9:21






  • 1




    @DietrichBurde Similar question is OK, but Letingok asks what's wrong with his proof which seems to be 10 times shorter.
    – Oldboy
    Nov 18 at 9:30






  • 3




    why should $x^3$ divide $24xyz$ ?
    – mercio
    Nov 18 at 9:35


















  • @ Dietrich Burde ,but in the begining we assume $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:21










  • $(1)$ should be $(a+b-c)^3-3(a+b)(c-a)(c-b)=c^3-a^3-b^3$.
    – Dietrich Burde
    Nov 18 at 9:21












  • @DietrichBurde (1) is true assuming that $a^3+b^3=c^3$?
    – Oldboy
    Nov 18 at 9:21






  • 1




    @DietrichBurde Similar question is OK, but Letingok asks what's wrong with his proof which seems to be 10 times shorter.
    – Oldboy
    Nov 18 at 9:30






  • 3




    why should $x^3$ divide $24xyz$ ?
    – mercio
    Nov 18 at 9:35
















@ Dietrich Burde ,but in the begining we assume $a^3+b^3=c^3$
– Leitingok
Nov 18 at 9:21




@ Dietrich Burde ,but in the begining we assume $a^3+b^3=c^3$
– Leitingok
Nov 18 at 9:21












$(1)$ should be $(a+b-c)^3-3(a+b)(c-a)(c-b)=c^3-a^3-b^3$.
– Dietrich Burde
Nov 18 at 9:21






$(1)$ should be $(a+b-c)^3-3(a+b)(c-a)(c-b)=c^3-a^3-b^3$.
– Dietrich Burde
Nov 18 at 9:21














@DietrichBurde (1) is true assuming that $a^3+b^3=c^3$?
– Oldboy
Nov 18 at 9:21




@DietrichBurde (1) is true assuming that $a^3+b^3=c^3$?
– Oldboy
Nov 18 at 9:21




1




1




@DietrichBurde Similar question is OK, but Letingok asks what's wrong with his proof which seems to be 10 times shorter.
– Oldboy
Nov 18 at 9:30




@DietrichBurde Similar question is OK, but Letingok asks what's wrong with his proof which seems to be 10 times shorter.
– Oldboy
Nov 18 at 9:30




3




3




why should $x^3$ divide $24xyz$ ?
– mercio
Nov 18 at 9:35




why should $x^3$ divide $24xyz$ ?
– mercio
Nov 18 at 9:35










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Your main result says that:



$$(x-y-z)^3=24xyz$$



...whcih basically means that $P=24xyz$ must be a perfect cube. Fair enough, but why does it implicate that $x^3$ divides $P$?



For example, $P$ is a perfect cube for $x=9$, $y=10^3$, $z=11^3$. But $9^3nmid24times9times10^3times11^3$.






share|cite|improve this answer



















  • 1




    this result in this post is by the assumption $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:40












  • It's easy to see that $x$ divides $c^3$, $y$ devides $a^3$,$z$ dvides $b^3$. $a^3,b^3,c^3$ are pairwise coprime,if $x,y,z$ are not pairwise coprime, for example $(x,y)=d$, then $d$ is common factor of $a^3$ and $c^3$, contradction.
    – Leitingok
    Nov 18 at 9:48













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Your main result says that:



$$(x-y-z)^3=24xyz$$



...whcih basically means that $P=24xyz$ must be a perfect cube. Fair enough, but why does it implicate that $x^3$ divides $P$?



For example, $P$ is a perfect cube for $x=9$, $y=10^3$, $z=11^3$. But $9^3nmid24times9times10^3times11^3$.






share|cite|improve this answer



















  • 1




    this result in this post is by the assumption $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:40












  • It's easy to see that $x$ divides $c^3$, $y$ devides $a^3$,$z$ dvides $b^3$. $a^3,b^3,c^3$ are pairwise coprime,if $x,y,z$ are not pairwise coprime, for example $(x,y)=d$, then $d$ is common factor of $a^3$ and $c^3$, contradction.
    – Leitingok
    Nov 18 at 9:48

















up vote
1
down vote



accepted










Your main result says that:



$$(x-y-z)^3=24xyz$$



...whcih basically means that $P=24xyz$ must be a perfect cube. Fair enough, but why does it implicate that $x^3$ divides $P$?



For example, $P$ is a perfect cube for $x=9$, $y=10^3$, $z=11^3$. But $9^3nmid24times9times10^3times11^3$.






share|cite|improve this answer



















  • 1




    this result in this post is by the assumption $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:40












  • It's easy to see that $x$ divides $c^3$, $y$ devides $a^3$,$z$ dvides $b^3$. $a^3,b^3,c^3$ are pairwise coprime,if $x,y,z$ are not pairwise coprime, for example $(x,y)=d$, then $d$ is common factor of $a^3$ and $c^3$, contradction.
    – Leitingok
    Nov 18 at 9:48















up vote
1
down vote



accepted







up vote
1
down vote



accepted






Your main result says that:



$$(x-y-z)^3=24xyz$$



...whcih basically means that $P=24xyz$ must be a perfect cube. Fair enough, but why does it implicate that $x^3$ divides $P$?



For example, $P$ is a perfect cube for $x=9$, $y=10^3$, $z=11^3$. But $9^3nmid24times9times10^3times11^3$.






share|cite|improve this answer














Your main result says that:



$$(x-y-z)^3=24xyz$$



...whcih basically means that $P=24xyz$ must be a perfect cube. Fair enough, but why does it implicate that $x^3$ divides $P$?



For example, $P$ is a perfect cube for $x=9$, $y=10^3$, $z=11^3$. But $9^3nmid24times9times10^3times11^3$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 18 at 10:43

























answered Nov 18 at 9:36









Oldboy

6,0081628




6,0081628








  • 1




    this result in this post is by the assumption $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:40












  • It's easy to see that $x$ divides $c^3$, $y$ devides $a^3$,$z$ dvides $b^3$. $a^3,b^3,c^3$ are pairwise coprime,if $x,y,z$ are not pairwise coprime, for example $(x,y)=d$, then $d$ is common factor of $a^3$ and $c^3$, contradction.
    – Leitingok
    Nov 18 at 9:48
















  • 1




    this result in this post is by the assumption $a^3+b^3=c^3$
    – Leitingok
    Nov 18 at 9:40












  • It's easy to see that $x$ divides $c^3$, $y$ devides $a^3$,$z$ dvides $b^3$. $a^3,b^3,c^3$ are pairwise coprime,if $x,y,z$ are not pairwise coprime, for example $(x,y)=d$, then $d$ is common factor of $a^3$ and $c^3$, contradction.
    – Leitingok
    Nov 18 at 9:48










1




1




this result in this post is by the assumption $a^3+b^3=c^3$
– Leitingok
Nov 18 at 9:40






this result in this post is by the assumption $a^3+b^3=c^3$
– Leitingok
Nov 18 at 9:40














It's easy to see that $x$ divides $c^3$, $y$ devides $a^3$,$z$ dvides $b^3$. $a^3,b^3,c^3$ are pairwise coprime,if $x,y,z$ are not pairwise coprime, for example $(x,y)=d$, then $d$ is common factor of $a^3$ and $c^3$, contradction.
– Leitingok
Nov 18 at 9:48






It's easy to see that $x$ divides $c^3$, $y$ devides $a^3$,$z$ dvides $b^3$. $a^3,b^3,c^3$ are pairwise coprime,if $x,y,z$ are not pairwise coprime, for example $(x,y)=d$, then $d$ is common factor of $a^3$ and $c^3$, contradction.
– Leitingok
Nov 18 at 9:48




















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