I tried to solve Fermat's last theorem! A legendary conundrum that remained unsolved for 350 years. by 天風 20231106

axa+bxb=cxc has an answer of 3×3+4×4=5×5, for example. However, Fermat's argument was that expressions with powers greater than axaxa+bxbxb=cxcxc, axaxaxa+bxbxbxb=cxcxcxc, and more do not have integer answers.

``It is impossible to divide a cubic number into two cubic numbers, or a 4th power into two 4th powers, and in general into the same power for any higher power than the 2nd power. I have discovered a truly amazing proof, but this space is too small to contain it.'' (The First Number Theory, Joseph H. Silverman)

In 1994, Andrew Wiles published a proof that all semistable elliptic curves with rational coefficients are modular. As a result, Fermat's prediction made 350 years ago became a reality.

(*) From here, I would like to write down the proof that I performed myself.

a^n+b^n=c^n (here, a^n means a is multiplied n times. If a^3, it is the same as axaxa) b^n=c^na^n

(Set as a=cd)

b^n=c^n-(cd)^n

=c^nc^n+du (Here, set (cd)^n=c^n-du)

b^n= du

If b=d then b^(n-1)=u

but d^(n-1)≠u

As a hint, a^3+b^3=(a+b)(a^2-ab+b^2), so (a^2-ab+b^2)≠(a+b)^2. Similarly, a^4+b^4 and a^5+b^5 do not become c^4 or c^5.