Does anyone know what is the status of solutions of Fermat's equation x^n+y^n=z^n for x,y,z other numbers such as 1) integers, 2) algebraic numbers, specially Q[i] and Q(i), complex numbers?
In which numbers does the equation have solutions?
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$\begingroup$ I can promise you that this is open. Fermat's Last Theorem was barely proven. $\endgroup$– Rushabh MehtaCommented Nov 16, 2019 at 7:35
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$\begingroup$ Well, some of these are not open - the algebraic numbers are an algebraically closed field, so any polynomial has the correct number of roots. Integers reduces to the case of natural numbers: if $n$ is even, just flip the signs on the negative integers, if $n$ is odd, just move the negatives integers to the other side. The other case of $\Bbb Q[i]=\Bbb Q(i)$ I do not immediately know the answer to, though. $\endgroup$– KReiserCommented Nov 16, 2019 at 7:59
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1 Answer
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Here are some solutions for all the cases asked:
$2^3+(-2)^3=0^3$ (for integers)
$1^4+0^4=i^4$ (complex)
$(2^{\frac{1}{3}})^3+(5^{\frac{1}{3}})^3=(7^{\frac{1}{3}})^3$ (algebraic)
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$\begingroup$ The Gaussian integers may be more interesting if $0$ is not allowed. $\endgroup$– badjohnCommented Nov 16, 2019 at 9:59
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$\begingroup$ I am sorry I should have realised Q[i]=Q(i) and that complex and algebraic cases are not applicable. $\endgroup$ Commented Nov 16, 2019 at 10:56