フェルマーの最終定理を解いちゃった
フェルマーの最終定理
3以上の自然数nについて
x^n+y^n=z^n
となる自然数の組(x,y,z)は存在しない
n=3の時
x^3+y^3=z^3
x^3+y^3=αとおくと
z^3=α
β1=(1*α)^(1/4)
β2=(β1*α)^(1/4)
z≒βm=(βm-1*α)^(1/4)
m=3の時
z≒β3=(((((1*α)^(1/4))*α)^(1/4))*α)^(1/4)
=((((α^(1/4))*α)^(1/4))*α)^(1/4)
zが整数であるためにはα^(1/4)が整数つまり
(x^3+y^3)^(1/4)≒zでxyzが整数である必要だから
x^3+y^3≒z^4とも取れこれと
x^3+y^3=z^3と同時に満たす整数xyzは存在しない
以下同様
証明終了
んーちょっと待ったよく考える
n=2の時
x^2+y^2=z^2
ピタゴラス数はm>nで
a=m^2-n^2,b=2mn=α,c=m^2+n^2
α=b=y=(z^2-x^2)^(1/2)=(c^2-a^2)^(1/2)
=((m^2+n^2)^2-(m^2-n^2)^2)^(1/2)
=((m^4+2m^2n^2+n^4)+(-m^4+2m^2n^2-n^4))^(1/2)
=(4m^2n^2)^(1/2)
=2mn
(c+a)/2=m^2,(c-a)/2=n^2
なので
例えばm=2,n=1で
a,b,c=3,4,5等
α=x^2=z^2-y^2
β1=(1*α)^(1/3)
β2=(β1*α)^(1/3)
x≒βm=(βm-1*α)^(1/3)=(β1^3)/x=α/x
だからこうかな
n=3の時
x^3+y^3=z^3
3次のピタゴラス数はm>nで
a^3+b^3=c^3
a=m^3-n^3,b=α,c=m^3+n^3
(c+a)/2=m^3,(c-a)/2=n^3
b=α=(c^3-a^3)^(1/3)
=((m^3+n^3)^3-(m^3-n^3)^3)^(1/3)
=((m^9+3m^6n^3+3m^3n^6+n^9)+(-m^9+3m^6n^3-3m^3n^6+n^9))^(1/3)
=(6m^6n^3+2n^9)^(1/3)
b=(6m^6n^3+2n^9)^(1/3)=(c^3-a^3)^(1/3)
b^3=(6m^6n^3+2n^9)=(c^3-a^3)
b^3=6m^6n^3+2n^9
b^3=n^9(6m^6/n^6+2)
b^3=n^9(6m^6/n^3+2^(1/2))(6m^6/n^3-2^(1/2))
b^3=αn^9(34)=αn^9(2*17)
bが自然数となるような自然数m>nは
α=(2*2*17*17)=1156
b^3=(1156)(34)=39304
b=34
b^3=6m^6n^3+2n^9
0=-39304+6m^6n^3+2n^9
2n^3=p
0=-39304+6m^3p+(1/2)p^3
(39304-(1/2)p^3)/6p=m^3
m=(((39304/6p)-(1/12)p^2))^(1/3)
c+a=2((39304/6p)-(1/12)p^2)
a^3+b^3=c^3
b^3=c^3-a^3=c^3-(2((39304/6p)-(1/12)p^2)-c)^3
34=-2((39304/6p)+(1/12)p^2)
0=-34-(39304/6)n^3+(2/3)n^6
(n^3+d)^2=0
d=((39304/6)±((39304/6)^2-(4)(2/3)(34))^(1/2))/(4/3)
d=((6550.666...)±((42,911,233.777...)-(90.666...))^(1/2))/(4/3)
d=((6550.666...)±(42911143.111...)^(1/2))/(4/3)
d=((6550.666...)±(6550.659...))/(4/3)
d1=117468.434...
n=(117468.434...)^(1/3)=48.975...
0=-39304+6m^6n^3+2n^9
0=-39304+6m^6(117468.434...)+2(117468.434...)^3
m^6=-(3241854603843501.475...)/6(117468.434...)
m=(-4599610995.415...)-(1/6)=40.781...
m=40.781...,n=48.975...
mn交換
(c+a)=2m^3,
(c+a)=2(48.975)^3=234936.868...
(c-a)=2(40.781...)^3=67822.471...
c=(234936.868...+67822.471...)/2=151379.669...
a=(151379.669...)-(67822.471...)=83557.198...
確かめ算
a^3+b^3c^3
83557.198...^3+=151379.669...^3
583380108791033.039...+39304=5.833e14=3.468e15
あーー全然ダメだったよい
やっぱり四次の近似でやってみる
n=2の時
x^2+y^2=z^2
x^2+y^2=αとおくと
z^2=α
β1=(1*α)^(1/3)
β2=(β1*α)^(1/3)
z≒βm=(βm-1*α)^(1/3)
α^(1/3)≒(x^2+y^2)^(1/3)=(x^2+y^2)/z=z
だから例えばx,y,z=3,4,5
n=3の時
x^3+y^3=z^3
x^3+y^3=αとおくと
z^3=α
β1=(1*α)^(1/4)
β2=(β1*α)^(1/4)
z≒βm=(βm-1*α)^(1/4)
α^(1/4)≒(x^3+y^3)^(1/4)=(x^3+y^3)/z^2=z
(x^3+y^3)/z^2=z
(x^3+y^3)/((x^3+y^3)/z^2)^2=z
(x^3+y^3)/((x^3+y^3)^2/z^4)=z
z^4(x^3+y^3)/(x^3+y^3)^2=z
z^4/(x^3+y^3)=z
z^4/z^3=z
z≒βm=(βm-1*α)^(1/4)
βm^4/z^3
(βm-1*(x^3+y^3))/(x^3+y^3)
(βm-2*(x^3+y^3))^(1/4)
≒((x^3+y^3))^(1/4)*(x^3+y^3))^(1/4))
=(x^3+y^3))^(1/2))=z^(3/2)≠z
以下同様
証明終了
戻る