Mixing
Fermat's Last Theorem solved through induction on $z$, not on $n$
$begingroup$
Assume that we have shown a proof for the case $z=1$ and $z=2$ for $x^n +y^n = z^n$.
Can we say that via mathematical induction on $z$ that this is now true for all integers $z$ for $n>2?$
number-theory
edited Jan 16 at 15:26
Mohammad Zuhair Khan
1,5852625
asked Jan 16 at 14:28
azdinazdin
34
$endgroup$
add a comment |
$begingroup$
Assume that we have shown a proof for the case $z=1$ and $z=2$ for $x^n +y^n = z^n$.
Can we say that via mathematical induction on $z$ that this is now true for all integers $z$ for $n>2?$
number-theory
edited Jan 16 at 15:26
Mohammad Zuhair Khan
1,5852625
asked Jan 16 at 14:28
azdinazdin
34
$endgroup$
2
$begingroup$
How would you do the induction step? Clearly the base case $x^n+y^n=1$ is easy.
$endgroup$
– Dietrich Burde
Jan 16 at 14:36
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Can you prove it for $z=1?$
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 14:39
$begingroup$
We can safely assume that if Fermat's last theorem could be solved via induction on $z$ , the theorem would have been proven long before Andrew Wiles's proof. Concluding from $z=1$ and $z=2$ to all $z$ is of course not possible : Simple example : $1$ and $2$ are divisors of $4$. Does that mean that all positive integers are divisors of $4$ ? Of course, no.
$endgroup$
– Peter
Jan 16 at 23:24
add a comment |
$begingroup$
Assume that we have shown a proof for the case $z=1$ and $z=2$ for $x^n +y^n = z^n$.
Can we say that via mathematical induction on $z$ that this is now true for all integers $z$ for $n>2?$
number-theory
edited Jan 16 at 15:26
Mohammad Zuhair Khan
1,5852625
asked Jan 16 at 14:28
azdinazdin
34
$endgroup$
Assume that we have shown a proof for the case $z=1$ and $z=2$ for $x^n +y^n = z^n$.
Can we say that via mathematical induction on $z$ that this is now true for all integers $z$ for $n>2?$
number-theory
number-theory
edited Jan 16 at 15:26
Mohammad Zuhair Khan
1,5852625
asked Jan 16 at 14:28
azdinazdin
34
edited Jan 16 at 15:26
Mohammad Zuhair Khan
1,5852625
asked Jan 16 at 14:28
azdinazdin
34
edited Jan 16 at 15:26
Mohammad Zuhair Khan
1,5852625
edited Jan 16 at 15:26
Mohammad Zuhair Khan
1,5852625
edited Jan 16 at 15:26
Mohammad Zuhair Khan
1,5852625
1,5852625
asked Jan 16 at 14:28
azdinazdin
34
asked Jan 16 at 14:28
azdinazdin
34
asked Jan 16 at 14:28
azdinazdin
34
34
2
$begingroup$
How would you do the induction step? Clearly the base case $x^n+y^n=1$ is easy.
$endgroup$
– Dietrich Burde
Jan 16 at 14:36
$begingroup$
Can you prove it for $z=1?$
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 14:39
$begingroup$
We can safely assume that if Fermat's last theorem could be solved via induction on $z$ , the theorem would have been proven long before Andrew Wiles's proof. Concluding from $z=1$ and $z=2$ to all $z$ is of course not possible : Simple example : $1$ and $2$ are divisors of $4$. Does that mean that all positive integers are divisors of $4$ ? Of course, no.
$endgroup$
– Peter
Jan 16 at 23:24
add a comment |
2
$begingroup$
How would you do the induction step? Clearly the base case $x^n+y^n=1$ is easy.
$endgroup$
– Dietrich Burde
Jan 16 at 14:36
$begingroup$
Can you prove it for $z=1?$
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 14:39
$begingroup$
We can safely assume that if Fermat's last theorem could be solved via induction on $z$ , the theorem would have been proven long before Andrew Wiles's proof. Concluding from $z=1$ and $z=2$ to all $z$ is of course not possible : Simple example : $1$ and $2$ are divisors of $4$. Does that mean that all positive integers are divisors of $4$ ? Of course, no.
$endgroup$
– Peter
Jan 16 at 23:24
2
2
$begingroup$
How would you do the induction step? Clearly the base case $x^n+y^n=1$ is easy.
$endgroup$
– Dietrich Burde
Jan 16 at 14:36
$begingroup$
How would you do the induction step? Clearly the base case $x^n+y^n=1$ is easy.
$endgroup$
– Dietrich Burde
Jan 16 at 14:36
$begingroup$
Can you prove it for $z=1?$
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 14:39
$begingroup$
Can you prove it for $z=1?$
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 14:39
$begingroup$
We can safely assume that if Fermat's last theorem could be solved via induction on $z$ , the theorem would have been proven long before Andrew Wiles's proof. Concluding from $z=1$ and $z=2$ to all $z$ is of course not possible : Simple example : $1$ and $2$ are divisors of $4$. Does that mean that all positive integers are divisors of $4$ ? Of course, no.
$endgroup$
– Peter
Jan 16 at 23:24
$begingroup$
We can safely assume that if Fermat's last theorem could be solved via induction on $z$ , the theorem would have been proven long before Andrew Wiles's proof. Concluding from $z=1$ and $z=2$ to all $z$ is of course not possible : Simple example : $1$ and $2$ are divisors of $4$. Does that mean that all positive integers are divisors of $4$ ? Of course, no.
$endgroup$
– Peter
Jan 16 at 23:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'm afraid this isn't quite how proof by induction works.
Call $P$ whatever we're proving about $z$. We can't simply say "$P$ is true for $z=1$ and $z=2$, therefore it's true for all $z>2$". (For example, "$z<3$" is true for both of them but false for $z>2$.)
To prove $P$ by induction we have to show that $P$ being true for one case forces it to be true for the next. Then we say 'Well it's true for $z=1$ so it must be true for $z=2$, so it must be true for $z=3$, so . . . " and let this count its way through all the remaining positive integers. But we've not got infinitely long in which to do the counting, so we compress that into " so by induction, it's true for all $zgeq 1$".
The key point is that each case must prove the next one, so there's a chain all the way up from a case that we already know to be true.
So I'm afraid your proof needs to do rather more.
answered Jan 16 at 15:36
timtfjtimtfj
2,448420
$endgroup$
$begingroup$
+1, although "this isn't quite how proof by induction works" is putting it rather mildly ...
$endgroup$
– Noah Schweber
Jan 16 at 15:37
$begingroup$
@NoahSchweber Ithink "do rather more" is mild too given how much effort Andrew Wiles had to put in!
$endgroup$
– timtfj
Jan 16 at 15:40
add a comment |
$begingroup$
Mathematical induction isn't "It works for $1$, it works for $2$, so it always works." You have to prove that the statement in question being true for $k$ implies that it is also true for $k+1$ - for any value of $k$ (or an equivalent variant of this).
What you seem to be doing is assuming that the induction step is given as proved already, which isn't the case - usually in fact this is precisely the nontrivial part of a proof by induction.
So no, you cannot say anything about FLT.
answered Jan 16 at 15:37
Noah SchweberNoah Schweber
125k10150287
$endgroup$
$begingroup$
Case z=1 gives 1^n +1^n /= 1^n , 1^n+2^n /=1^n, 2^n+3^n /= 1^n FLT is true for z=1 whatever the value for n and x, y integers. This is the basis step.
$endgroup$
– azdin
Jan 16 at 23:10
$begingroup$
@azdin And the induction step is ... what, precisely? Wishful thinking?
$endgroup$
– Noah Schweber
Jan 16 at 23:45
$begingroup$
Induction step: We assume FLT is true until z=(p-1) and we must proove that FLT is true for z=p. The case z=(p-1) implies the case z=p.
$endgroup$
– azdin
Jan 17 at 9:06
$begingroup$
@azdin "... and we must proove ..." yes, you must prove. Having not actually done that, you cannot say that FLT is proved. "The case z=(p-1) implies the case z=p." - how do you know that?
$endgroup$
– Noah Schweber
Jan 17 at 16:18
$begingroup$
I can say that FLT is proved because I have done it . I demonstrate that the case z=(p-1) implies z=p . I will send the proof in the next post in order to be reviewed by mathematicians and followers.
$endgroup$
– azdin
Jan 17 at 23:25
|
show 4 more comments
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm afraid this isn't quite how proof by induction works.
Call $P$ whatever we're proving about $z$. We can't simply say "$P$ is true for $z=1$ and $z=2$, therefore it's true for all $z>2$". (For example, "$z<3$" is true for both of them but false for $z>2$.)
To prove $P$ by induction we have to show that $P$ being true for one case forces it to be true for the next. Then we say 'Well it's true for $z=1$ so it must be true for $z=2$, so it must be true for $z=3$, so . . . " and let this count its way through all the remaining positive integers. But we've not got infinitely long in which to do the counting, so we compress that into " so by induction, it's true for all $zgeq 1$".
The key point is that each case must prove the next one, so there's a chain all the way up from a case that we already know to be true.
So I'm afraid your proof needs to do rather more.
answered Jan 16 at 15:36
timtfjtimtfj
2,448420
$endgroup$
$begingroup$
+1, although "this isn't quite how proof by induction works" is putting it rather mildly ...
$endgroup$
– Noah Schweber
Jan 16 at 15:37
$begingroup$
@NoahSchweber Ithink "do rather more" is mild too given how much effort Andrew Wiles had to put in!
$endgroup$
– timtfj
Jan 16 at 15:40
add a comment |
$begingroup$
I'm afraid this isn't quite how proof by induction works.
Call $P$ whatever we're proving about $z$. We can't simply say "$P$ is true for $z=1$ and $z=2$, therefore it's true for all $z>2$". (For example, "$z<3$" is true for both of them but false for $z>2$.)
To prove $P$ by induction we have to show that $P$ being true for one case forces it to be true for the next. Then we say 'Well it's true for $z=1$ so it must be true for $z=2$, so it must be true for $z=3$, so . . . " and let this count its way through all the remaining positive integers. But we've not got infinitely long in which to do the counting, so we compress that into " so by induction, it's true for all $zgeq 1$".
The key point is that each case must prove the next one, so there's a chain all the way up from a case that we already know to be true.
So I'm afraid your proof needs to do rather more.
answered Jan 16 at 15:36
timtfjtimtfj
2,448420
$endgroup$
$begingroup$
+1, although "this isn't quite how proof by induction works" is putting it rather mildly ...
$endgroup$
– Noah Schweber
Jan 16 at 15:37
$begingroup$
@NoahSchweber Ithink "do rather more" is mild too given how much effort Andrew Wiles had to put in!
$endgroup$
– timtfj
Jan 16 at 15:40
add a comment |
$begingroup$
I'm afraid this isn't quite how proof by induction works.
Call $P$ whatever we're proving about $z$. We can't simply say "$P$ is true for $z=1$ and $z=2$, therefore it's true for all $z>2$". (For example, "$z<3$" is true for both of them but false for $z>2$.)
To prove $P$ by induction we have to show that $P$ being true for one case forces it to be true for the next. Then we say 'Well it's true for $z=1$ so it must be true for $z=2$, so it must be true for $z=3$, so . . . " and let this count its way through all the remaining positive integers. But we've not got infinitely long in which to do the counting, so we compress that into " so by induction, it's true for all $zgeq 1$".
The key point is that each case must prove the next one, so there's a chain all the way up from a case that we already know to be true.
So I'm afraid your proof needs to do rather more.
answered Jan 16 at 15:36
timtfjtimtfj
2,448420
$endgroup$
I'm afraid this isn't quite how proof by induction works.
Call $P$ whatever we're proving about $z$. We can't simply say "$P$ is true for $z=1$ and $z=2$, therefore it's true for all $z>2$". (For example, "$z<3$" is true for both of them but false for $z>2$.)
To prove $P$ by induction we have to show that $P$ being true for one case forces it to be true for the next. Then we say 'Well it's true for $z=1$ so it must be true for $z=2$, so it must be true for $z=3$, so . . . " and let this count its way through all the remaining positive integers. But we've not got infinitely long in which to do the counting, so we compress that into " so by induction, it's true for all $zgeq 1$".
The key point is that each case must prove the next one, so there's a chain all the way up from a case that we already know to be true.
So I'm afraid your proof needs to do rather more.
answered Jan 16 at 15:36
timtfjtimtfj
2,448420
answered Jan 16 at 15:36
timtfjtimtfj
2,448420
answered Jan 16 at 15:36
timtfjtimtfj
2,448420
answered Jan 16 at 15:36
timtfjtimtfj
2,448420
2,448420
$begingroup$
+1, although "this isn't quite how proof by induction works" is putting it rather mildly ...
$endgroup$
– Noah Schweber
Jan 16 at 15:37
$begingroup$
@NoahSchweber Ithink "do rather more" is mild too given how much effort Andrew Wiles had to put in!
$endgroup$
– timtfj
Jan 16 at 15:40
add a comment |
$begingroup$
+1, although "this isn't quite how proof by induction works" is putting it rather mildly ...
$endgroup$
– Noah Schweber
Jan 16 at 15:37
$begingroup$
@NoahSchweber Ithink "do rather more" is mild too given how much effort Andrew Wiles had to put in!
$endgroup$
– timtfj
Jan 16 at 15:40
$begingroup$
+1, although "this isn't quite how proof by induction works" is putting it rather mildly ...
$endgroup$
– Noah Schweber
Jan 16 at 15:37
$begingroup$
+1, although "this isn't quite how proof by induction works" is putting it rather mildly ...
$endgroup$
– Noah Schweber
Jan 16 at 15:37
$begingroup$
@NoahSchweber Ithink "do rather more" is mild too given how much effort Andrew Wiles had to put in!
$endgroup$
– timtfj
Jan 16 at 15:40
$begingroup$
@NoahSchweber Ithink "do rather more" is mild too given how much effort Andrew Wiles had to put in!
$endgroup$
– timtfj
Jan 16 at 15:40
add a comment |
$begingroup$
Mathematical induction isn't "It works for $1$, it works for $2$, so it always works." You have to prove that the statement in question being true for $k$ implies that it is also true for $k+1$ - for any value of $k$ (or an equivalent variant of this).
What you seem to be doing is assuming that the induction step is given as proved already, which isn't the case - usually in fact this is precisely the nontrivial part of a proof by induction.
So no, you cannot say anything about FLT.
answered Jan 16 at 15:37
Noah SchweberNoah Schweber
125k10150287
$endgroup$
$begingroup$
Case z=1 gives 1^n +1^n /= 1^n , 1^n+2^n /=1^n, 2^n+3^n /= 1^n FLT is true for z=1 whatever the value for n and x, y integers. This is the basis step.
$endgroup$
– azdin
Jan 16 at 23:10
$begingroup$
@azdin And the induction step is ... what, precisely? Wishful thinking?
$endgroup$
– Noah Schweber
Jan 16 at 23:45
$begingroup$
Induction step: We assume FLT is true until z=(p-1) and we must proove that FLT is true for z=p. The case z=(p-1) implies the case z=p.
$endgroup$
– azdin
Jan 17 at 9:06
$begingroup$
@azdin "... and we must proove ..." yes, you must prove. Having not actually done that, you cannot say that FLT is proved. "The case z=(p-1) implies the case z=p." - how do you know that?
$endgroup$
– Noah Schweber
Jan 17 at 16:18
$begingroup$
I can say that FLT is proved because I have done it . I demonstrate that the case z=(p-1) implies z=p . I will send the proof in the next post in order to be reviewed by mathematicians and followers.
$endgroup$
– azdin
Jan 17 at 23:25
|
show 4 more comments
$begingroup$
Mathematical induction isn't "It works for $1$, it works for $2$, so it always works." You have to prove that the statement in question being true for $k$ implies that it is also true for $k+1$ - for any value of $k$ (or an equivalent variant of this).
What you seem to be doing is assuming that the induction step is given as proved already, which isn't the case - usually in fact this is precisely the nontrivial part of a proof by induction.
So no, you cannot say anything about FLT.
answered Jan 16 at 15:37
Noah SchweberNoah Schweber
125k10150287
$endgroup$
$begingroup$
Case z=1 gives 1^n +1^n /= 1^n , 1^n+2^n /=1^n, 2^n+3^n /= 1^n FLT is true for z=1 whatever the value for n and x, y integers. This is the basis step.
$endgroup$
– azdin
Jan 16 at 23:10
$begingroup$
@azdin And the induction step is ... what, precisely? Wishful thinking?
$endgroup$
– Noah Schweber
Jan 16 at 23:45
$begingroup$
Induction step: We assume FLT is true until z=(p-1) and we must proove that FLT is true for z=p. The case z=(p-1) implies the case z=p.
$endgroup$
– azdin
Jan 17 at 9:06
$begingroup$
@azdin "... and we must proove ..." yes, you must prove. Having not actually done that, you cannot say that FLT is proved. "The case z=(p-1) implies the case z=p." - how do you know that?
$endgroup$
– Noah Schweber
Jan 17 at 16:18
$begingroup$
I can say that FLT is proved because I have done it . I demonstrate that the case z=(p-1) implies z=p . I will send the proof in the next post in order to be reviewed by mathematicians and followers.
$endgroup$
– azdin
Jan 17 at 23:25
|
show 4 more comments
$begingroup$
Mathematical induction isn't "It works for $1$, it works for $2$, so it always works." You have to prove that the statement in question being true for $k$ implies that it is also true for $k+1$ - for any value of $k$ (or an equivalent variant of this).
What you seem to be doing is assuming that the induction step is given as proved already, which isn't the case - usually in fact this is precisely the nontrivial part of a proof by induction.
So no, you cannot say anything about FLT.
answered Jan 16 at 15:37
Noah SchweberNoah Schweber
125k10150287
$endgroup$
Mathematical induction isn't "It works for $1$, it works for $2$, so it always works." You have to prove that the statement in question being true for $k$ implies that it is also true for $k+1$ - for any value of $k$ (or an equivalent variant of this).
What you seem to be doing is assuming that the induction step is given as proved already, which isn't the case - usually in fact this is precisely the nontrivial part of a proof by induction.
So no, you cannot say anything about FLT.
answered Jan 16 at 15:37
Noah SchweberNoah Schweber
125k10150287
answered Jan 16 at 15:37
Noah SchweberNoah Schweber
125k10150287
answered Jan 16 at 15:37
Noah SchweberNoah Schweber
125k10150287
answered Jan 16 at 15:37
Noah SchweberNoah Schweber
125k10150287
125k10150287
$begingroup$
Case z=1 gives 1^n +1^n /= 1^n , 1^n+2^n /=1^n, 2^n+3^n /= 1^n FLT is true for z=1 whatever the value for n and x, y integers. This is the basis step.
$endgroup$
– azdin
Jan 16 at 23:10
$begingroup$
@azdin And the induction step is ... what, precisely? Wishful thinking?
$endgroup$
– Noah Schweber
Jan 16 at 23:45
$begingroup$
Induction step: We assume FLT is true until z=(p-1) and we must proove that FLT is true for z=p. The case z=(p-1) implies the case z=p.
$endgroup$
– azdin
Jan 17 at 9:06
$begingroup$
@azdin "... and we must proove ..." yes, you must prove. Having not actually done that, you cannot say that FLT is proved. "The case z=(p-1) implies the case z=p." - how do you know that?
$endgroup$
– Noah Schweber
Jan 17 at 16:18
$begingroup$
I can say that FLT is proved because I have done it . I demonstrate that the case z=(p-1) implies z=p . I will send the proof in the next post in order to be reviewed by mathematicians and followers.
$endgroup$
– azdin
Jan 17 at 23:25
|
show 4 more comments
$begingroup$
Case z=1 gives 1^n +1^n /= 1^n , 1^n+2^n /=1^n, 2^n+3^n /= 1^n FLT is true for z=1 whatever the value for n and x, y integers. This is the basis step.
$endgroup$
– azdin
Jan 16 at 23:10
$begingroup$
@azdin And the induction step is ... what, precisely? Wishful thinking?
$endgroup$
– Noah Schweber
Jan 16 at 23:45
$begingroup$
Induction step: We assume FLT is true until z=(p-1) and we must proove that FLT is true for z=p. The case z=(p-1) implies the case z=p.
$endgroup$
– azdin
Jan 17 at 9:06
$begingroup$
@azdin "... and we must proove ..." yes, you must prove. Having not actually done that, you cannot say that FLT is proved. "The case z=(p-1) implies the case z=p." - how do you know that?
$endgroup$
– Noah Schweber
Jan 17 at 16:18
$begingroup$
I can say that FLT is proved because I have done it . I demonstrate that the case z=(p-1) implies z=p . I will send the proof in the next post in order to be reviewed by mathematicians and followers.
$endgroup$
– azdin
Jan 17 at 23:25
$begingroup$
Case z=1 gives 1^n +1^n /= 1^n , 1^n+2^n /=1^n, 2^n+3^n /= 1^n FLT is true for z=1 whatever the value for n and x, y integers. This is the basis step.
$endgroup$
– azdin
Jan 16 at 23:10
$begingroup$
Case z=1 gives 1^n +1^n /= 1^n , 1^n+2^n /=1^n, 2^n+3^n /= 1^n FLT is true for z=1 whatever the value for n and x, y integers. This is the basis step.
$endgroup$
– azdin
Jan 16 at 23:10
$begingroup$
@azdin And the induction step is ... what, precisely? Wishful thinking?
$endgroup$
– Noah Schweber
Jan 16 at 23:45
$begingroup$
@azdin And the induction step is ... what, precisely? Wishful thinking?
$endgroup$
– Noah Schweber
Jan 16 at 23:45
$begingroup$
Induction step: We assume FLT is true until z=(p-1) and we must proove that FLT is true for z=p. The case z=(p-1) implies the case z=p.
$endgroup$
– azdin
Jan 17 at 9:06
$begingroup$
Induction step: We assume FLT is true until z=(p-1) and we must proove that FLT is true for z=p. The case z=(p-1) implies the case z=p.
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– azdin
Jan 17 at 9:06
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@azdin "... and we must proove ..." yes, you must prove. Having not actually done that, you cannot say that FLT is proved. "The case z=(p-1) implies the case z=p." - how do you know that?
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– Noah Schweber
Jan 17 at 16:18
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@azdin "... and we must proove ..." yes, you must prove. Having not actually done that, you cannot say that FLT is proved. "The case z=(p-1) implies the case z=p." - how do you know that?
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– Noah Schweber
Jan 17 at 16:18
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I can say that FLT is proved because I have done it . I demonstrate that the case z=(p-1) implies z=p . I will send the proof in the next post in order to be reviewed by mathematicians and followers.
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– azdin
Jan 17 at 23:25
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I can say that FLT is proved because I have done it . I demonstrate that the case z=(p-1) implies z=p . I will send the proof in the next post in order to be reviewed by mathematicians and followers.
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– azdin
Jan 17 at 23:25
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How would you do the induction step? Clearly the base case $x^n+y^n=1$ is easy.
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– Dietrich Burde
Jan 16 at 14:36
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Can you prove it for $z=1?$
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– Mohammad Zuhair Khan
Jan 16 at 14:39
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We can safely assume that if Fermat's last theorem could be solved via induction on $z$ , the theorem would have been proven long before Andrew Wiles's proof. Concluding from $z=1$ and $z=2$ to all $z$ is of course not possible : Simple example : $1$ and $2$ are divisors of $4$. Does that mean that all positive integers are divisors of $4$ ? Of course, no.
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– Peter
Jan 16 at 23:24