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A proof of Fermat´s Last Theorem using only Gauss´s Lemma for the roots of monic polynomials with integer coefficients.**

Writing the Fermat equation $$a^n + b^n - c^n = (c-p)^n + (c-q)^n - c^n = 0$$ with $q < p$ being integers and c taken as an independent variable we get for $n=3$ the polynomial equation $$F_3(c,p,q) = c^3 - 3(p + q)c^2 + 3(p^2+q^2)c -(p^3+q^3) = 0.$$ In order to preserve the parity always two of $p, q, c$ must be odd and one even. So for an odd $c$ one of $p$ or $q$ must be odd and the other even (first case), whereas for an even $c$ both $p$ and $q$ must be odd (second case). Therefore in both cases $F_3(c,p,q)$ has an even value and might be zero because zero is an even number.

In order to show that this never can be we compare in the first case the roots of $F_3(c,p,q)=0$ to the roots of $F_3(c,2p,2q)=0$, which for an odd $c$ is odd and therefore cannot be zero. Hence $F_3(c,2p,2q)=0$ according to Gauss´s Lemma must have irrational roots which according to Vieta´s formulas for the roots and coefficients of polynomials are twice the roots of $F_3(c,p,q)=0$, which therefore are also irrational. Thus $F_3(c,p,q)$ in the first case never can be zero for an odd $c$.

In the second case $c$ is even and can be written as $2^t*d$, where $d$ is the odd part of $c$ and $t$ is an integer. Rewriting $F_3(c,p,q)$ as $$F_3(2^td, 2^tp,2^tq) = 2^3t [d^3 – 3 (p + q)d^2 + 3 (p^2+q^2)d -(p^3+q^3)] = 0$$ which never can be zero because of the odd value of the expression in square brackets.

With the substitution of $p$ and $q$ by $2^tp$ and $2^tq$ the roots of $F_3(2^td,2^tp,2^tq)=0$ are $2^t$-fold the roots of $F_3(c,p,q)=0$ and are irrational according to Gauss´s Lemma. Thus the roots of $F_3(c,p,q)=0$ are also irrational which shows that it in the second case also never can be zero for an even $c$.

Therefore the Fermat equation $a^3 + b^3 - c^3 = 0$ for $n=3$ never can be zero and thus cannot have a solution in integers $a,b,c$. This result can easily be extended to all degrees $n$ being odd primes using the features of the binomial coefficients, especially the fact that for all $n$ there is always an even number of odd coefficients (see, please, https://www.researchgate.net/publication/338819565_Proving_Fermats_Last_Theorem_as_a_Mean_Value_Problem_Title_Proving_Fermats_Last_Theorem_as_a_Mean_Value_Problem_Author).

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    $\begingroup$ here is a good tutorial on Mathjax. $\endgroup$ – lulu May 23 at 10:10
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    $\begingroup$ What do you mean by "the root" of a polynomial in three variables? $\endgroup$ – joriki May 23 at 19:41
  • $\begingroup$ There is only one variable, namely c, and the cubic polynomial equation must have three roots. $\endgroup$ – Joachim Gantz May 24 at 9:11
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    $\begingroup$ I think people might be more willing to comb through your proof if you improve the formatting, as it’s currently a pretty unappealing wall of text. Some paragraphs and single line equations would go a long way. $\endgroup$ – Qwertiops May 24 at 11:06
  • $\begingroup$ If you consider only $c$ a variable and not $p$ and $q$, I would recommend making that explicit in the notation and/or in the text. Currently I don't see how I could have deduced it from either. Also, the fact that the cubic polynomial equation must have three roots doesn't help me understand what you mean by "the root". In case English is not your native language: The use of the definite article "the" in "the root" implies that there exists exactly one root, to which this expression refers. $\endgroup$ – joriki May 24 at 11:36
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The flaw seems to be in reasoning that if $F_3(x,2p,2q)$ for odd $x=c$ is odd, and so cannot have rational root eventually, then $F_3(x,p,q)$ could not have a rational root for $x=c$ odd. But, $c$ was supposed to be odd in $F_3(x,p,q)$ with $x=c$, while in $F_3(x,2p,2q)$ we expect a root $x=2c$, which is even. You have incorrectly expected that the $F_3(x,2p,2q)$ would have to have its root odd as well.

If you still don't see the flaw, notice that your proof would show that the original equation had no solution, but that is not true. There are known trivial solutions, one is corresponding to $c=1,p=1,q=0$. Here $F_3(x,p,q)=(x-1)^3$ (an odd root), but $F_3(x,2p,2q)=(x-2)^3$ (an even root). Another missed solution is $p=q=c=0$. Try to go through your argument with these specific values and you will see the issue yourself.

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  • $\begingroup$ The relation between the roots of F3(x,2p,2q)=0 and F3(x,p,q)=0 can be seen if F3(x,2p,2q)=0 is divided by $2^3$. Then it becomes F3(x/2,p,q)=0 which shows that the roots of F3(x,p,q)=0 for the same p and q are exactly half of the roots of F3(x,2p,2q)=0 . As F3(x,2p,2q)=0 for x= an odd c has irrational roots, so the roots of F3(x,p,q)=0 must be also irrational. $\endgroup$ – Joachim Gantz Jun 2 at 9:00
  • $\begingroup$ Maximum you can show is that odd $c$ cannot be root of $F_3(x,2p,2q)$, but that does not mean its roots are irrational (as in example of $F_3(x,2p,2q)=(x-2)^3$ with $p=1,q=0$, which does not have odd root, yet it has rational root). Overall looking for odd $c$ in $F_3(x,2p,2q)$ is irrelevant (it does not lead to the conclusion, and $c$ is a root of $F_3(x,p,q)$, then $2c$ is a root of $F_3(x,2p,2q)$, and $2c$ is even, so not useful there either). $\endgroup$ – Sil Jun 2 at 9:22
  • $\begingroup$ What do you think about the following numerical example with p=2 and q=1? F3(x,2p,2q) =0 gives x1=14.1091..., whereas F3(x,p,q)=0 gives 7.05455..., exactly half of the former. F3(c,2p,2q) cannot have an integer root, because for an odd x=c=7 it has an odd value which cannot be zero. So x=7 is not a root. Besides, this example corresponds to $5^3+6^3 = 7^3 (-2).$ $\endgroup$ – Joachim Gantz Jun 2 at 11:18
  • $\begingroup$ That example works because $F_3(x,2p,2q)$ has no rational root, therefore $F_3(x,p,q)$ has no rat root. But it is not true that $F_3(x,2p,2q)$ has no rational root BECAUSE it has no odd root, that does not follow. You would have to rule out possibility of even root as well (which conincidently there is none in this case). For some reason you think that $F_3(x,2p,2q)$ can have only odd roots if any, but as I explained before, if the $c$ was odd in $F_3(x,p,q)$, then the correspoding root in $F_3(x,2p,2q)$ will be $2c$. So if anything, you would need to show that $F_3(x,2p,2q)$ has no even root. $\endgroup$ – Sil Jun 2 at 11:48
  • $\begingroup$ Really try to write out the example with $p=1,q=0$ and see that argumentation that "$F_3(x,2p,2q)$ has no odd root and so it has no rational root" does not work (because exactly in that case it has no odd root but has even root). $\endgroup$ – Sil Jun 2 at 11:52

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