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Is this a proof of Fermat's Last Theorem where n = 3?
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So I have spent some time thinking about Fermat's Last Theorem and about how to come up with a proof for certain cases of n. To begin with I took n = 3.
$$A^3 + B^3 = C^3 (A,B,C in mathbb Z)( A,B,C ≠0)$$
We can assume that $A<B<C$.
Because of this we can rewrite this as:
$$A^3 + xA^3 = C^3$$
We now have three possibilities:
$x$ is an integer, $x$ is an irrational, or $x$ is a fraction.
If $x$ is an integer then $x$+1 and $x$ must be cubes. The only numbers which satisfy this are $x$ = 0. However, this means $B^3 = 0$.
$x$ cannot be irrational because then $C^3$ is irrational.
Therefore $x$ is a fraction.
We can represent this now as:
$$A^3 + fracpqB^3 = C^3 (p,q in mathbb Z)(GCD(p,q) = 1)$$
This means $fracpq$ is actually a cube over a cube. Therefore $p$ and $q$ are cube numbers.
However, $1 + fracpq$ is also a cube. Therefore $fracp+qq$ is a cube over a cube. Hence: $$sqrt[3]p ,sqrt[3]q, sqrt[3]p+q in mathbb Z$$
We can now create another variable $v$ for $p+q$.
Therefore, $$p+q = v$$ (Where $sqrt[3]v in mathbb Z$)
Now we must deal with $fracpq+1$. Since this is a cube we can write this as:
$$fracpq + 1 = lambda^3$$
Since $p$ and $q$ are co-prime and in simplest form, $lambda$ and $lambda^3$ are also fractions in simplest form. Rearranging, we get:
$$p = q(lambda^3 - 1)$$
Substituting this into $p+q=v$, we get:
$$q+q(lambda^3 - 1) = v$$
$$qlambda^3 = v$$
Since $lambda^3$ is a fraction, we can express it as $fracD^3E^3$.
Hence:
$$qD^3 = vE^3$$
However, because $v$ is a cube, $vE^3$ is also a cube. Therefore $lambda^3$ doesn't have to be a fraction. This is a contradiction, and so there can be no integer solutions for $p$ and $q$. Subsequently, there can be no integer solutions for A,B,C in:
$$A^3 + B^3 = C^3$$.
If anyone could point out the problem in this and what you think of it it would be greatly appreciated.
Thank you.
number-theory
asked Jul 22 at 12:18
roskiller
1267
 |Â
show 2 more comments
up vote
0
down vote
favorite
So I have spent some time thinking about Fermat's Last Theorem and about how to come up with a proof for certain cases of n. To begin with I took n = 3.
$$A^3 + B^3 = C^3 (A,B,C in mathbb Z)( A,B,C ≠0)$$
We can assume that $A<B<C$.
Because of this we can rewrite this as:
$$A^3 + xA^3 = C^3$$
We now have three possibilities:
$x$ is an integer, $x$ is an irrational, or $x$ is a fraction.
If $x$ is an integer then $x$+1 and $x$ must be cubes. The only numbers which satisfy this are $x$ = 0. However, this means $B^3 = 0$.
$x$ cannot be irrational because then $C^3$ is irrational.
Therefore $x$ is a fraction.
We can represent this now as:
$$A^3 + fracpqB^3 = C^3 (p,q in mathbb Z)(GCD(p,q) = 1)$$
This means $fracpq$ is actually a cube over a cube. Therefore $p$ and $q$ are cube numbers.
However, $1 + fracpq$ is also a cube. Therefore $fracp+qq$ is a cube over a cube. Hence: $$sqrt[3]p ,sqrt[3]q, sqrt[3]p+q in mathbb Z$$
We can now create another variable $v$ for $p+q$.
Therefore, $$p+q = v$$ (Where $sqrt[3]v in mathbb Z$)
Now we must deal with $fracpq+1$. Since this is a cube we can write this as:
$$fracpq + 1 = lambda^3$$
Since $p$ and $q$ are co-prime and in simplest form, $lambda$ and $lambda^3$ are also fractions in simplest form. Rearranging, we get:
$$p = q(lambda^3 - 1)$$
Substituting this into $p+q=v$, we get:
$$q+q(lambda^3 - 1) = v$$
$$qlambda^3 = v$$
Since $lambda^3$ is a fraction, we can express it as $fracD^3E^3$.
Hence:
$$qD^3 = vE^3$$
However, because $v$ is a cube, $vE^3$ is also a cube. Therefore $lambda^3$ doesn't have to be a fraction. This is a contradiction, and so there can be no integer solutions for $p$ and $q$. Subsequently, there can be no integer solutions for A,B,C in:
$$A^3 + B^3 = C^3$$.
If anyone could point out the problem in this and what you think of it it would be greatly appreciated.
Thank you.
number-theory
asked Jul 22 at 12:18
roskiller
1267
How do you justify the sentence “This means $fracpq$ is actually a cube over a cubeâ€�
– José Carlos Santos
Jul 22 at 12:24
2
The very definition of $x$ implies that $x=fracB^3A^3$
– random
Jul 22 at 12:27
2
"Therefore $lambda^3$ doesn't have to be a fraction" ? Formulated this way, it could be a fraction, so no contradiction arises. If you mean "$lambda^3$ cannot be a fraction", then justify why this is the case !
– Peter
Jul 22 at 12:30
Why is "$1+frac pq$ also a cube"?
– Arnaud Mortier
Jul 22 at 12:55
2
After the conclusion that $x$ is a fraction, shouldn't the next formula be $A^3+frac p q A^3 = C^3$ ?
– random
Jul 22 at 13:14
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I have spent some time thinking about Fermat's Last Theorem and about how to come up with a proof for certain cases of n. To begin with I took n = 3.
$$A^3 + B^3 = C^3 (A,B,C in mathbb Z)( A,B,C ≠0)$$
We can assume that $A<B<C$.
Because of this we can rewrite this as:
$$A^3 + xA^3 = C^3$$
We now have three possibilities:
$x$ is an integer, $x$ is an irrational, or $x$ is a fraction.
If $x$ is an integer then $x$+1 and $x$ must be cubes. The only numbers which satisfy this are $x$ = 0. However, this means $B^3 = 0$.
$x$ cannot be irrational because then $C^3$ is irrational.
Therefore $x$ is a fraction.
We can represent this now as:
$$A^3 + fracpqB^3 = C^3 (p,q in mathbb Z)(GCD(p,q) = 1)$$
This means $fracpq$ is actually a cube over a cube. Therefore $p$ and $q$ are cube numbers.
However, $1 + fracpq$ is also a cube. Therefore $fracp+qq$ is a cube over a cube. Hence: $$sqrt[3]p ,sqrt[3]q, sqrt[3]p+q in mathbb Z$$
We can now create another variable $v$ for $p+q$.
Therefore, $$p+q = v$$ (Where $sqrt[3]v in mathbb Z$)
Now we must deal with $fracpq+1$. Since this is a cube we can write this as:
$$fracpq + 1 = lambda^3$$
Since $p$ and $q$ are co-prime and in simplest form, $lambda$ and $lambda^3$ are also fractions in simplest form. Rearranging, we get:
$$p = q(lambda^3 - 1)$$
Substituting this into $p+q=v$, we get:
$$q+q(lambda^3 - 1) = v$$
$$qlambda^3 = v$$
Since $lambda^3$ is a fraction, we can express it as $fracD^3E^3$.
Hence:
$$qD^3 = vE^3$$
However, because $v$ is a cube, $vE^3$ is also a cube. Therefore $lambda^3$ doesn't have to be a fraction. This is a contradiction, and so there can be no integer solutions for $p$ and $q$. Subsequently, there can be no integer solutions for A,B,C in:
$$A^3 + B^3 = C^3$$.
If anyone could point out the problem in this and what you think of it it would be greatly appreciated.
Thank you.
number-theory
asked Jul 22 at 12:18
roskiller
1267
So I have spent some time thinking about Fermat's Last Theorem and about how to come up with a proof for certain cases of n. To begin with I took n = 3.
$$A^3 + B^3 = C^3 (A,B,C in mathbb Z)( A,B,C ≠0)$$
We can assume that $A<B<C$.
Because of this we can rewrite this as:
$$A^3 + xA^3 = C^3$$
We now have three possibilities:
$x$ is an integer, $x$ is an irrational, or $x$ is a fraction.
If $x$ is an integer then $x$+1 and $x$ must be cubes. The only numbers which satisfy this are $x$ = 0. However, this means $B^3 = 0$.
$x$ cannot be irrational because then $C^3$ is irrational.
Therefore $x$ is a fraction.
We can represent this now as:
$$A^3 + fracpqB^3 = C^3 (p,q in mathbb Z)(GCD(p,q) = 1)$$
This means $fracpq$ is actually a cube over a cube. Therefore $p$ and $q$ are cube numbers.
However, $1 + fracpq$ is also a cube. Therefore $fracp+qq$ is a cube over a cube. Hence: $$sqrt[3]p ,sqrt[3]q, sqrt[3]p+q in mathbb Z$$
We can now create another variable $v$ for $p+q$.
Therefore, $$p+q = v$$ (Where $sqrt[3]v in mathbb Z$)
Now we must deal with $fracpq+1$. Since this is a cube we can write this as:
$$fracpq + 1 = lambda^3$$
Since $p$ and $q$ are co-prime and in simplest form, $lambda$ and $lambda^3$ are also fractions in simplest form. Rearranging, we get:
$$p = q(lambda^3 - 1)$$
Substituting this into $p+q=v$, we get:
$$q+q(lambda^3 - 1) = v$$
$$qlambda^3 = v$$
Since $lambda^3$ is a fraction, we can express it as $fracD^3E^3$.
Hence:
$$qD^3 = vE^3$$
However, because $v$ is a cube, $vE^3$ is also a cube. Therefore $lambda^3$ doesn't have to be a fraction. This is a contradiction, and so there can be no integer solutions for $p$ and $q$. Subsequently, there can be no integer solutions for A,B,C in:
$$A^3 + B^3 = C^3$$.
If anyone could point out the problem in this and what you think of it it would be greatly appreciated.
Thank you.
number-theory
asked Jul 22 at 12:18
roskiller
1267
asked Jul 22 at 12:18
roskiller
1267
asked Jul 22 at 12:18
roskiller
1267
asked Jul 22 at 12:18
roskiller
1267
1267
How do you justify the sentence “This means $fracpq$ is actually a cube over a cubeâ€�
– José Carlos Santos
Jul 22 at 12:24
2
The very definition of $x$ implies that $x=fracB^3A^3$
– random
Jul 22 at 12:27
2
"Therefore $lambda^3$ doesn't have to be a fraction" ? Formulated this way, it could be a fraction, so no contradiction arises. If you mean "$lambda^3$ cannot be a fraction", then justify why this is the case !
– Peter
Jul 22 at 12:30
Why is "$1+frac pq$ also a cube"?
– Arnaud Mortier
Jul 22 at 12:55
2
After the conclusion that $x$ is a fraction, shouldn't the next formula be $A^3+frac p q A^3 = C^3$ ?
– random
Jul 22 at 13:14
 |Â
show 2 more comments
How do you justify the sentence “This means $fracpq$ is actually a cube over a cubeâ€�
– José Carlos Santos
Jul 22 at 12:24
2
The very definition of $x$ implies that $x=fracB^3A^3$
– random
Jul 22 at 12:27
2
"Therefore $lambda^3$ doesn't have to be a fraction" ? Formulated this way, it could be a fraction, so no contradiction arises. If you mean "$lambda^3$ cannot be a fraction", then justify why this is the case !
– Peter
Jul 22 at 12:30
Why is "$1+frac pq$ also a cube"?
– Arnaud Mortier
Jul 22 at 12:55
2
After the conclusion that $x$ is a fraction, shouldn't the next formula be $A^3+frac p q A^3 = C^3$ ?
– random
Jul 22 at 13:14
How do you justify the sentence “This means $fracpq$ is actually a cube over a cubeâ€�
– José Carlos Santos
Jul 22 at 12:24
How do you justify the sentence “This means $fracpq$ is actually a cube over a cubeâ€�
– José Carlos Santos
Jul 22 at 12:24
2
2
The very definition of $x$ implies that $x=fracB^3A^3$
– random
Jul 22 at 12:27
The very definition of $x$ implies that $x=fracB^3A^3$
– random
Jul 22 at 12:27
2
2
"Therefore $lambda^3$ doesn't have to be a fraction" ? Formulated this way, it could be a fraction, so no contradiction arises. If you mean "$lambda^3$ cannot be a fraction", then justify why this is the case !
– Peter
Jul 22 at 12:30
"Therefore $lambda^3$ doesn't have to be a fraction" ? Formulated this way, it could be a fraction, so no contradiction arises. If you mean "$lambda^3$ cannot be a fraction", then justify why this is the case !
– Peter
Jul 22 at 12:30
Why is "$1+frac pq$ also a cube"?
– Arnaud Mortier
Jul 22 at 12:55
Why is "$1+frac pq$ also a cube"?
– Arnaud Mortier
Jul 22 at 12:55
2
2
After the conclusion that $x$ is a fraction, shouldn't the next formula be $A^3+frac p q A^3 = C^3$ ?
– random
Jul 22 at 13:14
After the conclusion that $x$ is a fraction, shouldn't the next formula be $A^3+frac p q A^3 = C^3$ ?
– random
Jul 22 at 13:14
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
You have
$$qD^3=v E^3.tag1$$
But if you've followed carefully, you'd have noticed that
$D=C$, $E=B$, $p=A^3$, $q=B^3$ and $v=C^3$. The conclusion given is
then that "$lambda^3$ doesn't have to be a fraction". I presume
this is intended to mean that $D^3/E^3notinBbb Z$. But (1) says
that $B^3C^3=C^3B^3$. From this tautology, why can we deduce anything
non-trivial at all, let alone $D^3/E^3notinBbb Z$?
answered Jul 22 at 13:21
Lord Shark the Unknown
85.2k950111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You have
$$qD^3=v E^3.tag1$$
But if you've followed carefully, you'd have noticed that
$D=C$, $E=B$, $p=A^3$, $q=B^3$ and $v=C^3$. The conclusion given is
then that "$lambda^3$ doesn't have to be a fraction". I presume
this is intended to mean that $D^3/E^3notinBbb Z$. But (1) says
that $B^3C^3=C^3B^3$. From this tautology, why can we deduce anything
non-trivial at all, let alone $D^3/E^3notinBbb Z$?
answered Jul 22 at 13:21
Lord Shark the Unknown
85.2k950111
add a comment |Â
up vote
3
down vote
accepted
You have
$$qD^3=v E^3.tag1$$
But if you've followed carefully, you'd have noticed that
$D=C$, $E=B$, $p=A^3$, $q=B^3$ and $v=C^3$. The conclusion given is
then that "$lambda^3$ doesn't have to be a fraction". I presume
this is intended to mean that $D^3/E^3notinBbb Z$. But (1) says
that $B^3C^3=C^3B^3$. From this tautology, why can we deduce anything
non-trivial at all, let alone $D^3/E^3notinBbb Z$?
answered Jul 22 at 13:21
Lord Shark the Unknown
85.2k950111
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You have
$$qD^3=v E^3.tag1$$
But if you've followed carefully, you'd have noticed that
$D=C$, $E=B$, $p=A^3$, $q=B^3$ and $v=C^3$. The conclusion given is
then that "$lambda^3$ doesn't have to be a fraction". I presume
this is intended to mean that $D^3/E^3notinBbb Z$. But (1) says
that $B^3C^3=C^3B^3$. From this tautology, why can we deduce anything
non-trivial at all, let alone $D^3/E^3notinBbb Z$?
answered Jul 22 at 13:21
Lord Shark the Unknown
85.2k950111
You have
$$qD^3=v E^3.tag1$$
But if you've followed carefully, you'd have noticed that
$D=C$, $E=B$, $p=A^3$, $q=B^3$ and $v=C^3$. The conclusion given is
then that "$lambda^3$ doesn't have to be a fraction". I presume
this is intended to mean that $D^3/E^3notinBbb Z$. But (1) says
that $B^3C^3=C^3B^3$. From this tautology, why can we deduce anything
non-trivial at all, let alone $D^3/E^3notinBbb Z$?
answered Jul 22 at 13:21
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 13:21
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 13:21
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 13:21
Lord Shark the Unknown
85.2k950111
85.2k950111
add a comment |Â
add a comment |Â
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How do you justify the sentence “This means $fracpq$ is actually a cube over a cubeâ€�
– José Carlos Santos
Jul 22 at 12:24
2
The very definition of $x$ implies that $x=fracB^3A^3$
– random
Jul 22 at 12:27
2
"Therefore $lambda^3$ doesn't have to be a fraction" ? Formulated this way, it could be a fraction, so no contradiction arises. If you mean "$lambda^3$ cannot be a fraction", then justify why this is the case !
– Peter
Jul 22 at 12:30
Why is "$1+frac pq$ also a cube"?
– Arnaud Mortier
Jul 22 at 12:55
2
After the conclusion that $x$ is a fraction, shouldn't the next formula be $A^3+frac p q A^3 = C^3$ ?
– random
Jul 22 at 13:14