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.
elementary-number-theory
asked Nov 18 at 9:06
Leitingok
941932
|
show 2 more comments
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.
elementary-number-theory
asked Nov 18 at 9:06
Leitingok
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
|
show 2 more comments
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.
elementary-number-theory
asked Nov 18 at 9:06
Leitingok
941932
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
elementary-number-theory
asked Nov 18 at 9:06
Leitingok
941932
asked Nov 18 at 9:06
Leitingok
941932
edited Nov 18 at 9:15
asked Nov 18 at 9:06
Leitingok
941932
asked Nov 18 at 9:06
Leitingok
941932
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
|
show 2 more comments
@ 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
|
show 2 more comments
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 $9^3nmid24times9times10^3times11^3$.
answered Nov 18 at 9:36
Oldboy
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
add a comment |
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 $9^3nmid24times9times10^3times11^3$.
answered Nov 18 at 9:36
Oldboy
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
add a comment |
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 $9^3nmid24times9times10^3times11^3$.
answered Nov 18 at 9:36
Oldboy
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
add a comment |
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 $9^3nmid24times9times10^3times11^3$.
answered Nov 18 at 9:36
Oldboy
6,0081628
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 $9^3nmid24times9times10^3times11^3$.
answered Nov 18 at 9:36
Oldboy
6,0081628
edited Nov 18 at 10:43
answered Nov 18 at 9:36
Oldboy
6,0081628
answered Nov 18 at 9:36
Oldboy
6,0081628
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
add a comment |
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
add a comment |
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@ 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