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I'm reading Fermat's Last Theorem by Simon Singh and in it he writes the following:

... if Fermat's Last Theorem turned out to be undecidable, then this would imply that it must be true. The reason is as follows. The Last Theorem says there are no whole number solutions to the equation x^n + y^n = z^n for n greater than 2. If the Last Theorem were in fact false, then it would be possible to prove this by identifying solution (a counter-example). Therefore the Last Theorem would be decidable. Being false would be inconsistent with being undecidable. However, if the Last Theorem were true, there would not necessarily be such an unequivocal way of proving it so.

So to prove that the Last Theorem is true it would be enough to prove that it is undecidable since it would prove there are no counter-examples.

The Problem

If proving the Last Theorem is undecidable proves the Last Theorem is true, wouldn't the Last Theorem be in fact decidable (at least in this very roundabout way)? I feel like there is a concept I'm missing here.

Griffin Staples
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    Singh is being a bit careless. First, it should be "formally undecidable". Second, he is missing "in the standard model". That is, if FLT (or Goldbach's Conjecture) were formally undecidable, then it would be true in the standard model. There would be models where FLT is false, but the meaning of "number" would not be the one we are used to. – Arturo Magidin Jul 31 '23 at 18:05
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    You always have to be careful with arguments like this to specify the theories in which things are being proved. There are two theories in this argument: the theory $T_1$ over which FLT is assumed to be independent and the theory $T_2$ in which this independence itself is provable. What we can conclude is that if ($T_1$ and $T_2$ are each "appropriate" and) $T_2$ proves "FLT is independent of $T_1$" then $T_2$ proves FLT. But this isn't a contradiction at all. In particular, we can't possibly have $T_1=T_2$ by Godel's second incompleteness theorem. – Noah Schweber Jul 31 '23 at 18:06
  • (I've left this as a comment as opposed to an answer because I'm quite certain this has been asked before on this site; I even recall answering this question ...) – Noah Schweber Jul 31 '23 at 18:06
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    There is no contradiction. If a stronger theory (say ZFC) can prove that FLT is undecidable in a weaker , but still strong enough theory (say PA) , then it has proven the statement. The weaker theory can of course not detect this undecidability itself (this would be a contradiction). Note that we do not always have semi-decidability. Collatz for example , it is not obvious how we should prove that some sequence diverges. – Peter Jul 31 '23 at 18:13
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    Related: one, two, three, four, five, six. – Arturo Magidin Jul 31 '23 at 18:16
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    @NoahSchweber Perhaps this? – Arturo Magidin Jul 31 '23 at 18:19
  • Thanks for all the answers guys. I'm still trying to wrap my head around it but the stronger/weaker theories argument seems to make sense. – Griffin Staples Jul 31 '23 at 19:49

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