When does Fermat's Last Theorem hold over finite fields?
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It is well-known that in his attempts to prove Fermat's Last Theorem (FLT) over $mathbb Z^+$, Schur came up with a result that has come to be known as Schur's Theorem, which implies that FLT fails over $mathbb Z_p$ for all but finitely many primes $p$. One might see this intuitively by taking any expression of the form $x^n+y^n-z^n$, with $x,y,z,ninmathbb Z^+$ with $ngeq 3$, then simply picking any prime factor of this number. Hence, we know that FLT does not, in general, hold in finite fields.
But surely there are cases where FLT does hold in interesting finite fields as well, other than just for certain choices of $p$ and considering $mathbb Z_p$. What are some interesting finite fields in which FLT holds?
abstract-algebra number-theory field-theory finite-fields
asked Dec 10 '18 at 9:14
YiFanYiFan
3,9741627
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add a comment |
$begingroup$
It is well-known that in his attempts to prove Fermat's Last Theorem (FLT) over $mathbb Z^+$, Schur came up with a result that has come to be known as Schur's Theorem, which implies that FLT fails over $mathbb Z_p$ for all but finitely many primes $p$. One might see this intuitively by taking any expression of the form $x^n+y^n-z^n$, with $x,y,z,ninmathbb Z^+$ with $ngeq 3$, then simply picking any prime factor of this number. Hence, we know that FLT does not, in general, hold in finite fields.
But surely there are cases where FLT does hold in interesting finite fields as well, other than just for certain choices of $p$ and considering $mathbb Z_p$. What are some interesting finite fields in which FLT holds?
abstract-algebra number-theory field-theory finite-fields
asked Dec 10 '18 at 9:14
YiFanYiFan
3,9741627
$endgroup$
1
$begingroup$
I am not sure I understand what you are looking for. FLT failing means that the equation has solutions. Once you have solutions mod $p$, you also have them in all fields of characteristic $p$, so picking other finite fields can only make it fail harder.
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– Tobias Kildetoft
Dec 10 '18 at 9:43
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I do not understand the question well either. In any case, the opposite seems to be true. For instance, in the case of the finite field $ mathbb{F}_3 $ and $ n $ odd, we have a non-trivial solution $ (2,1,0) $ to $ x^n + y^n = z^n $. In general $ p $-adic analysis (Hensel's lemma) will give you solutions for specific cases.
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– hellHound
Dec 10 '18 at 19:57
add a comment |
$begingroup$
It is well-known that in his attempts to prove Fermat's Last Theorem (FLT) over $mathbb Z^+$, Schur came up with a result that has come to be known as Schur's Theorem, which implies that FLT fails over $mathbb Z_p$ for all but finitely many primes $p$. One might see this intuitively by taking any expression of the form $x^n+y^n-z^n$, with $x,y,z,ninmathbb Z^+$ with $ngeq 3$, then simply picking any prime factor of this number. Hence, we know that FLT does not, in general, hold in finite fields.
But surely there are cases where FLT does hold in interesting finite fields as well, other than just for certain choices of $p$ and considering $mathbb Z_p$. What are some interesting finite fields in which FLT holds?
abstract-algebra number-theory field-theory finite-fields
asked Dec 10 '18 at 9:14
YiFanYiFan
3,9741627
$endgroup$
It is well-known that in his attempts to prove Fermat's Last Theorem (FLT) over $mathbb Z^+$, Schur came up with a result that has come to be known as Schur's Theorem, which implies that FLT fails over $mathbb Z_p$ for all but finitely many primes $p$. One might see this intuitively by taking any expression of the form $x^n+y^n-z^n$, with $x,y,z,ninmathbb Z^+$ with $ngeq 3$, then simply picking any prime factor of this number. Hence, we know that FLT does not, in general, hold in finite fields.
But surely there are cases where FLT does hold in interesting finite fields as well, other than just for certain choices of $p$ and considering $mathbb Z_p$. What are some interesting finite fields in which FLT holds?
abstract-algebra number-theory field-theory finite-fields
abstract-algebra number-theory field-theory finite-fields
asked Dec 10 '18 at 9:14
YiFanYiFan
3,9741627
asked Dec 10 '18 at 9:14
YiFanYiFan
3,9741627
asked Dec 10 '18 at 9:14
YiFanYiFan
3,9741627
asked Dec 10 '18 at 9:14
YiFanYiFan
3,9741627
asked Dec 10 '18 at 9:14
YiFanYiFan
3,9741627
3,9741627
1
$begingroup$
I am not sure I understand what you are looking for. FLT failing means that the equation has solutions. Once you have solutions mod $p$, you also have them in all fields of characteristic $p$, so picking other finite fields can only make it fail harder.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 9:43
$begingroup$
I do not understand the question well either. In any case, the opposite seems to be true. For instance, in the case of the finite field $ mathbb{F}_3 $ and $ n $ odd, we have a non-trivial solution $ (2,1,0) $ to $ x^n + y^n = z^n $. In general $ p $-adic analysis (Hensel's lemma) will give you solutions for specific cases.
$endgroup$
– hellHound
Dec 10 '18 at 19:57
add a comment |
1
$begingroup$
I am not sure I understand what you are looking for. FLT failing means that the equation has solutions. Once you have solutions mod $p$, you also have them in all fields of characteristic $p$, so picking other finite fields can only make it fail harder.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 9:43
$begingroup$
I do not understand the question well either. In any case, the opposite seems to be true. For instance, in the case of the finite field $ mathbb{F}_3 $ and $ n $ odd, we have a non-trivial solution $ (2,1,0) $ to $ x^n + y^n = z^n $. In general $ p $-adic analysis (Hensel's lemma) will give you solutions for specific cases.
$endgroup$
– hellHound
Dec 10 '18 at 19:57
1
1
$begingroup$
I am not sure I understand what you are looking for. FLT failing means that the equation has solutions. Once you have solutions mod $p$, you also have them in all fields of characteristic $p$, so picking other finite fields can only make it fail harder.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 9:43
$begingroup$
I am not sure I understand what you are looking for. FLT failing means that the equation has solutions. Once you have solutions mod $p$, you also have them in all fields of characteristic $p$, so picking other finite fields can only make it fail harder.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 9:43
$begingroup$
I do not understand the question well either. In any case, the opposite seems to be true. For instance, in the case of the finite field $ mathbb{F}_3 $ and $ n $ odd, we have a non-trivial solution $ (2,1,0) $ to $ x^n + y^n = z^n $. In general $ p $-adic analysis (Hensel's lemma) will give you solutions for specific cases.
$endgroup$
– hellHound
Dec 10 '18 at 19:57
$begingroup$
I do not understand the question well either. In any case, the opposite seems to be true. For instance, in the case of the finite field $ mathbb{F}_3 $ and $ n $ odd, we have a non-trivial solution $ (2,1,0) $ to $ x^n + y^n = z^n $. In general $ p $-adic analysis (Hensel's lemma) will give you solutions for specific cases.
$endgroup$
– hellHound
Dec 10 '18 at 19:57
add a comment |
1 Answer
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FLT does not hold over any finite field. Over $mathbb{F}_q$ we have $x^q = x$, so $x^q + y^q = z^q$ iff $x + y = z$. If $q = 2$ then take the exponent to be $4$.
answered Dec 11 '18 at 1:07
Qiaochu YuanQiaochu Yuan
279k32588932
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add a comment |
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$begingroup$
FLT does not hold over any finite field. Over $mathbb{F}_q$ we have $x^q = x$, so $x^q + y^q = z^q$ iff $x + y = z$. If $q = 2$ then take the exponent to be $4$.
answered Dec 11 '18 at 1:07
Qiaochu YuanQiaochu Yuan
279k32588932
$endgroup$
add a comment |
$begingroup$
FLT does not hold over any finite field. Over $mathbb{F}_q$ we have $x^q = x$, so $x^q + y^q = z^q$ iff $x + y = z$. If $q = 2$ then take the exponent to be $4$.
answered Dec 11 '18 at 1:07
Qiaochu YuanQiaochu Yuan
279k32588932
$endgroup$
add a comment |
$begingroup$
FLT does not hold over any finite field. Over $mathbb{F}_q$ we have $x^q = x$, so $x^q + y^q = z^q$ iff $x + y = z$. If $q = 2$ then take the exponent to be $4$.
answered Dec 11 '18 at 1:07
Qiaochu YuanQiaochu Yuan
279k32588932
$endgroup$
FLT does not hold over any finite field. Over $mathbb{F}_q$ we have $x^q = x$, so $x^q + y^q = z^q$ iff $x + y = z$. If $q = 2$ then take the exponent to be $4$.
answered Dec 11 '18 at 1:07
Qiaochu YuanQiaochu Yuan
279k32588932
answered Dec 11 '18 at 1:07
Qiaochu YuanQiaochu Yuan
279k32588932
answered Dec 11 '18 at 1:07
Qiaochu YuanQiaochu Yuan
279k32588932
answered Dec 11 '18 at 1:07
Qiaochu YuanQiaochu Yuan
279k32588932
279k32588932
add a comment |
add a comment |
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$begingroup$
I am not sure I understand what you are looking for. FLT failing means that the equation has solutions. Once you have solutions mod $p$, you also have them in all fields of characteristic $p$, so picking other finite fields can only make it fail harder.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 9:43
$begingroup$
I do not understand the question well either. In any case, the opposite seems to be true. For instance, in the case of the finite field $ mathbb{F}_3 $ and $ n $ odd, we have a non-trivial solution $ (2,1,0) $ to $ x^n + y^n = z^n $. In general $ p $-adic analysis (Hensel's lemma) will give you solutions for specific cases.
$endgroup$
– hellHound
Dec 10 '18 at 19:57