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Suppose that $\sqrt[3]{2}=\frac{p}{q}$. Then $2q^3 = p^3$ i.e $q^3 + q^3 = p^3$, which is contradiction with Fermat's Last Theorem.

My question is whether this argument is a correct mathematical proof, since Fermat's Last Theorem is proven, or does it loop on itself somewhere along the proof of the Theorem?

In other words, does the proof of Fermat's Theorem somehow rely on the fact that $\sqrt[3]{2}$ is irrational?

UPD: As pointed out in comments, this actually is a valid argument, no matter what was used in the proof of the Fermat's Last Theorem (which from now on will be referred to as the Proof). What really interests me, is whether the Proof uses on some step the fact that $\sqrt[3]{2}$ is irrational?

Glinka
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    Yes, but you don't need Fermat for this. If you have $2=p^3/q^3$ with coprime integers $q,p$, we must have $q^3=\pm 1$, hence $2=\pm p^3$, a contradiction. As to "in other words" - no, Fermat is much more general. – Dietrich Burde Nov 29 '15 at 12:28
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    Since the problem asked for a proof with Fermat's Last Theorem (a HUGE overkill), I think your solution has no flaws. – rkm0959 Nov 29 '15 at 12:29
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    If the big proof ever needed the fact that $\sqrt[3]2$ is irrational, it didn't take that for granted but somehow established it. So you are on the safe side. In other words, the Wile's theorem isn't phrased as "if $\sqrt[3]2$ is irrational, then $x^n+y^n=z^n$" has no solution, but is self-contained. –  Nov 29 '15 at 12:31
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    It is a proof as a matter of facts. FLT is a theorem, not a conjecture anymore. In any case, for cubic powers the argument holds also without FLT, because it can be proven that $x^3+y^3=z^3$ has no non-trivial integer solution using the method of infinite descent. – PITTALUGA Nov 29 '15 at 12:32
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    @DietrichBurde, of cause it is.This is just a fun observation, a way to fry a bacon using a rocket, and I what to know, whether it is correct. – Glinka Nov 29 '15 at 12:33
  • GyuminRoh, it's not a problem from a book or smth, I composed this title myself, so that's not an indicator – Glinka Nov 29 '15 at 12:35
  • If it uses a lemma somewhere that says "if $2p^3 = q^3$ then $|p| < 10$" (i.e. the lemma seem to almost say that it is irrational but doesnt really finish the job), would you say that the proof is circular ? – mercio Nov 29 '15 at 14:10

4 Answers4

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In this comment BCnrd argues that this proof is "essentially circular", because converting an FLT counterexample to a Frey curve with certain congruence conditions as in the Wiles proof requires an argument equivalent to establishing irrationality of $\sqrt[3]{2}$.

Community
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Matthew Towers
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Your argument is correct but there is no need to use Wiles' proof of Fermat's Last Theorem: an elementary proof of the case $n=3$ was given by Euler.

lhf
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  • Thanks, that's very useful comment! Why did I think I needed the general case... – Glinka Jun 01 '16 at 19:32
  • The day before Andrew Wiles published his corrected proof, FLT was an open question for irregular primes greater than $4,000,000$. So Wiles provided another method to prove that $\sqrt[n]2$ is irrational, if $n$ is an irregular prime and $n>4,000,000$ – Michael Ejercito Feb 29 '24 at 02:28
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In case not, I doubt anyone on Earth knows. The Wiles' proof is a huge document (150 pages), readable by only a few people, which indirectly involves the work of dozen (hundredths) mathematicians, thousand (million ?) pages of previous results. Unless you've read all this corpus, you can't tell whether the irrationality of $\sqrt[3]2$ is somewhere invoked or not.

In any case, there is no circular argument as the irrationality of $\sqrt[3]2$ can be established by a child of five.

user26857
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does the proof of Fermat's Theorem somehow rely on the fact that $\sqrt[3]{2}$ is irrational?

Edit Added: Even if $\sqrt[3]{2}$ is irrational was contained in FLT, it would have had to be proven by some means, so as long as FLT did not assume FLT then it doesn't matter that a specific instance of FLT was contained in proof of FLT

FLT being true implies $\sqrt[3]{2}$ is irrational , ($\sqrt[3]{2}$ is irrational is a specific case of FLT).

$\sqrt[3]{2}$ is irrational is not enough to imply FLT.

as $\sqrt[3]{2}$ is irrational is a specific instance of FLT, if it was not true then it would be in contradiction of a more general theorem containing it as a specific instance of the theorem.

in other words FLT implies all specific instances of itself, where as infinitely many instances of FLT do not imply FLT.

jimjim
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    No, that's not what the OP meant. He was asking whether the irrationality of $\sqrt[3]2$ is used anywhere in the proof of the FLT. Such a situation would mean that the OP's proof is circular, since it relies on the FLT, which in turn would rely on the irrationality of $\sqrt[3]2$. – Alex M. Nov 29 '15 at 13:48
  • @AlexM. even if $\sqrt[3]{2}$ is irrational was contained in FLT, it would have had to be proven by some means, so as long as FLT did not assume FLT then it doesn't matter that a specific instance of FLT was contained in proof of FLT – jimjim Nov 29 '15 at 13:55
  • @Arjang, yes, thank you, but that was already poited out. – Glinka Nov 29 '15 at 17:58
  • @Glinka : where? which comment ? – jimjim Dec 01 '15 at 13:07