Is this Diophantine problem solvable without invoking Fermat's Last Theorem?
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Let $a,b,c,n$ be positive integers with $a<b<c$ and $ngeq 3$ odd. Given that $a^n + b^n < 2c^n$, can one prove that $a^{n+2}+b^{n+2}neq c^{n+2}$ without invoking Wiles' theorem ? Or is this actually equivalent to Fermat's Last Theorem ?
diophantine-equations
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Let $a,b,c,n$ be positive integers with $a<b<c$ and $ngeq 3$ odd. Given that $a^n + b^n < 2c^n$, can one prove that $a^{n+2}+b^{n+2}neq c^{n+2}$ without invoking Wiles' theorem ? Or is this actually equivalent to Fermat's Last Theorem ?
diophantine-equations
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2
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If $a<b<c$ isn't $a^n+b^n<2c^n$ trivial? You see, $a<c$ implies that $a^n<c^n$ and $b<c$ implies that $b^n<c^n$.
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– Yanko
Dec 10 '18 at 15:44
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This seems trivially equivalent to Fermat. True, you exclude the case $a=b$ but $2a^n=c^n$ is clearly impossible (considering the powers of $2$ on both sides) so that's not a significant exclusion. Am I missing something?
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– lulu
Dec 10 '18 at 15:46
1
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@Yanko, ah, that's it, thanks !
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– user507152
Dec 10 '18 at 15:47
add a comment |
$begingroup$
Let $a,b,c,n$ be positive integers with $a<b<c$ and $ngeq 3$ odd. Given that $a^n + b^n < 2c^n$, can one prove that $a^{n+2}+b^{n+2}neq c^{n+2}$ without invoking Wiles' theorem ? Or is this actually equivalent to Fermat's Last Theorem ?
diophantine-equations
$endgroup$
Let $a,b,c,n$ be positive integers with $a<b<c$ and $ngeq 3$ odd. Given that $a^n + b^n < 2c^n$, can one prove that $a^{n+2}+b^{n+2}neq c^{n+2}$ without invoking Wiles' theorem ? Or is this actually equivalent to Fermat's Last Theorem ?
diophantine-equations
diophantine-equations
asked Dec 10 '18 at 15:39
user507152
2
$begingroup$
If $a<b<c$ isn't $a^n+b^n<2c^n$ trivial? You see, $a<c$ implies that $a^n<c^n$ and $b<c$ implies that $b^n<c^n$.
$endgroup$
– Yanko
Dec 10 '18 at 15:44
$begingroup$
This seems trivially equivalent to Fermat. True, you exclude the case $a=b$ but $2a^n=c^n$ is clearly impossible (considering the powers of $2$ on both sides) so that's not a significant exclusion. Am I missing something?
$endgroup$
– lulu
Dec 10 '18 at 15:46
1
$begingroup$
@Yanko, ah, that's it, thanks !
$endgroup$
– user507152
Dec 10 '18 at 15:47
add a comment |
2
$begingroup$
If $a<b<c$ isn't $a^n+b^n<2c^n$ trivial? You see, $a<c$ implies that $a^n<c^n$ and $b<c$ implies that $b^n<c^n$.
$endgroup$
– Yanko
Dec 10 '18 at 15:44
$begingroup$
This seems trivially equivalent to Fermat. True, you exclude the case $a=b$ but $2a^n=c^n$ is clearly impossible (considering the powers of $2$ on both sides) so that's not a significant exclusion. Am I missing something?
$endgroup$
– lulu
Dec 10 '18 at 15:46
1
$begingroup$
@Yanko, ah, that's it, thanks !
$endgroup$
– user507152
Dec 10 '18 at 15:47
2
2
$begingroup$
If $a<b<c$ isn't $a^n+b^n<2c^n$ trivial? You see, $a<c$ implies that $a^n<c^n$ and $b<c$ implies that $b^n<c^n$.
$endgroup$
– Yanko
Dec 10 '18 at 15:44
$begingroup$
If $a<b<c$ isn't $a^n+b^n<2c^n$ trivial? You see, $a<c$ implies that $a^n<c^n$ and $b<c$ implies that $b^n<c^n$.
$endgroup$
– Yanko
Dec 10 '18 at 15:44
$begingroup$
This seems trivially equivalent to Fermat. True, you exclude the case $a=b$ but $2a^n=c^n$ is clearly impossible (considering the powers of $2$ on both sides) so that's not a significant exclusion. Am I missing something?
$endgroup$
– lulu
Dec 10 '18 at 15:46
$begingroup$
This seems trivially equivalent to Fermat. True, you exclude the case $a=b$ but $2a^n=c^n$ is clearly impossible (considering the powers of $2$ on both sides) so that's not a significant exclusion. Am I missing something?
$endgroup$
– lulu
Dec 10 '18 at 15:46
1
1
$begingroup$
@Yanko, ah, that's it, thanks !
$endgroup$
– user507152
Dec 10 '18 at 15:47
$begingroup$
@Yanko, ah, that's it, thanks !
$endgroup$
– user507152
Dec 10 '18 at 15:47
add a comment |
1 Answer
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As stated in the comments, $a<b<c$ implies trivially $a^n+b^n<2c^n$, hence this condition is extraneous. The equation also immediately implies $a,b<c$ so that is not useful either. Finally the equation is symmetric in $a,b$ so we may WLOG assume $aleq b$, so your condition just becomes $aneq b$. In fact, $aneq b$ is quite extraneous too, because from the Fermat equation we will obtain $sqrt[n+2]{2}$ is rational if the equation has solutions with $a=b$. So what you're left with is, "prove that if $n$ is odd, then $a^{n+2}+b^{n+2}=c^{n+2}$ has no solutions." Now it is entirely obvious that this is just Fermat's Last Theorem in the odd case.
answered Dec 13 '18 at 13:26
YiFanYiFan
4,0021627
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$begingroup$
As stated in the comments, $a<b<c$ implies trivially $a^n+b^n<2c^n$, hence this condition is extraneous. The equation also immediately implies $a,b<c$ so that is not useful either. Finally the equation is symmetric in $a,b$ so we may WLOG assume $aleq b$, so your condition just becomes $aneq b$. In fact, $aneq b$ is quite extraneous too, because from the Fermat equation we will obtain $sqrt[n+2]{2}$ is rational if the equation has solutions with $a=b$. So what you're left with is, "prove that if $n$ is odd, then $a^{n+2}+b^{n+2}=c^{n+2}$ has no solutions." Now it is entirely obvious that this is just Fermat's Last Theorem in the odd case.
answered Dec 13 '18 at 13:26
YiFanYiFan
4,0021627
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add a comment |
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As stated in the comments, $a<b<c$ implies trivially $a^n+b^n<2c^n$, hence this condition is extraneous. The equation also immediately implies $a,b<c$ so that is not useful either. Finally the equation is symmetric in $a,b$ so we may WLOG assume $aleq b$, so your condition just becomes $aneq b$. In fact, $aneq b$ is quite extraneous too, because from the Fermat equation we will obtain $sqrt[n+2]{2}$ is rational if the equation has solutions with $a=b$. So what you're left with is, "prove that if $n$ is odd, then $a^{n+2}+b^{n+2}=c^{n+2}$ has no solutions." Now it is entirely obvious that this is just Fermat's Last Theorem in the odd case.
answered Dec 13 '18 at 13:26
YiFanYiFan
4,0021627
$endgroup$
add a comment |
$begingroup$
As stated in the comments, $a<b<c$ implies trivially $a^n+b^n<2c^n$, hence this condition is extraneous. The equation also immediately implies $a,b<c$ so that is not useful either. Finally the equation is symmetric in $a,b$ so we may WLOG assume $aleq b$, so your condition just becomes $aneq b$. In fact, $aneq b$ is quite extraneous too, because from the Fermat equation we will obtain $sqrt[n+2]{2}$ is rational if the equation has solutions with $a=b$. So what you're left with is, "prove that if $n$ is odd, then $a^{n+2}+b^{n+2}=c^{n+2}$ has no solutions." Now it is entirely obvious that this is just Fermat's Last Theorem in the odd case.
answered Dec 13 '18 at 13:26
YiFanYiFan
4,0021627
$endgroup$
As stated in the comments, $a<b<c$ implies trivially $a^n+b^n<2c^n$, hence this condition is extraneous. The equation also immediately implies $a,b<c$ so that is not useful either. Finally the equation is symmetric in $a,b$ so we may WLOG assume $aleq b$, so your condition just becomes $aneq b$. In fact, $aneq b$ is quite extraneous too, because from the Fermat equation we will obtain $sqrt[n+2]{2}$ is rational if the equation has solutions with $a=b$. So what you're left with is, "prove that if $n$ is odd, then $a^{n+2}+b^{n+2}=c^{n+2}$ has no solutions." Now it is entirely obvious that this is just Fermat's Last Theorem in the odd case.
answered Dec 13 '18 at 13:26
YiFanYiFan
4,0021627
answered Dec 13 '18 at 13:26
YiFanYiFan
4,0021627
answered Dec 13 '18 at 13:26
YiFanYiFan
4,0021627
answered Dec 13 '18 at 13:26
YiFanYiFan
4,0021627
4,0021627
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$begingroup$
If $a<b<c$ isn't $a^n+b^n<2c^n$ trivial? You see, $a<c$ implies that $a^n<c^n$ and $b<c$ implies that $b^n<c^n$.
$endgroup$
– Yanko
Dec 10 '18 at 15:44
$begingroup$
This seems trivially equivalent to Fermat. True, you exclude the case $a=b$ but $2a^n=c^n$ is clearly impossible (considering the powers of $2$ on both sides) so that's not a significant exclusion. Am I missing something?
$endgroup$
– lulu
Dec 10 '18 at 15:46
1
$begingroup$
@Yanko, ah, that's it, thanks !
$endgroup$
– user507152
Dec 10 '18 at 15:47