请输入您要查询的字词:

 

单词 ProposedElementaryProofOfFermatsLastTheorem
释义

proposed elementary proof of Fermat’s last theorem


Michael Pogorsky has offered what is said to be an elementary proof of Fermat’slast theoremMathworldPlanetmath. Is the proof correct? The intentof this entry is to show the proof up to the point at which it fails,if there is such a point. New equation numbers will be used.

proofWe assume that there are positive integers a,b and c such that

an+bn=cn   (1)

.

We can assume without loss of generality that a,b and c are mutually coprime, so that in factthey are also pairwise coprime. The proof is split into 3 major cases:(1) n is a prime greater than 2, (2) n is divisible by a prime greater than 2,and (3) n is a power of 2.

1 n is a prime greater than 2

Write c as

c=a+k=b+f

for some integers k and f.

Then

an+bn=(a+k)n=(b+f)n.

Using the binomial theorem we can write

an=f(nbn-1+12n(n-1)bn-2f++fn-1)   (2)

and

bn=k(nan-1+12n(n-1)an-2k++kn-1).   (3)

Lemma 1. If n is prime numberMathworldPlanetmath then n divides (nk) for 0<k<n.
The proof is easy.

Claim: gcd(f,k)=1.
Proof. A factor of f and k will also divide a and b by equations (2)and (3). But a and b are coprimeMathworldPlanetmath, so the gcd of f and k must be 1.

Now write (2) as an=fs for some integer s.

From that point in the proof the exposition is somewhat unclearso I will attempt to rearrange the steps in what seems to be a better order.First, I introduce a lemma of my own.Write (3) as bn=kt for some integer t.

Lemma 2. gcd(f,s)=nα and gcd(k,t)=nβ forsome nonnegative integers α and β.
Proof. Suppose q divides gcd(f,s) where q is a prime. Then q divides a and s.We can write s=nbn-1+fT for some integer T, so that q divides s-fT and thereforeq divides nbn-1. Hence q divides n or q divides bn-1. But if q dividesbn-1 then q divides b, a contradictionMathworldPlanetmathPlanetmath. Hence q divides n. But n is a prime, soq=n. From this we get that gcd(f,s)=nα for some nonnegative integer α.Similarly, gcd(k,t)=nβ forsome nonnegative integer β.

It is clear that at least one of α and β is zero, otherwise n divides fand k. Without loss of generality, we can assume that α=0.

The author now introduces what he calls version A and version B. I would prefer to call theseCase A and Case B. But there is no claim outstanding yet, so I have to defer the case split.What seems to be the next main result is stated in the following lemma.

Lemma 3. There exist positive integers p,u,v,w such
1) a=vp,
2) if β=0 then

a=uwv+vn
b=uwv+wn
c=uwv+vn+wn.

and
3) if β>0 then there is a positive integers g such that

a=nguwv+vn
b=nguwv+ngn-1wn

and

c=nguwv+vn+ngn-1wn.

Proof. (1) We have an=fs,where f and s are coprime. By unique factorization of integersit must be that f=vn and s=pn for some positive integers p and v. It follows thata=pv.
(2) Similarly, there are positive integers w and q such that b=wq, where wn=k.From a+k=b+f we get

vp+wn=wq+vn

and after regrouping we have

v(p-vn-1)=w(q-wn-1)

Since gcd(f,k)=1 it follows that gcd(v,w)=1, so that

v|(q-wn-1)

and

w|(p-vn-1).

Hence

u:=p-vn-1w=q-wn-1v   (4)

is an integer. Using u we can now writea=uwv+vn, b=uwv+wn, and c=uwv+vn+wn.
(3) Since α=0, we have gcd(f,s)=1.Since β>0, we can write

k=k1nτ

where τ>0 and gcd(k1,n)=1. By Lemma 1 there is a positive integer ci such that(ni)=nci for 0<i<n. We can write

t=i=1n(ni)an-iki-1=i=1n-1ncian-iki-1+kn-1
=i=1n-1ncian-iki-1+nk1nτ-1kn-2=nT

where T=an-1+12(n-1)an-2k++k1nτ-1kn-2.Hence,

bn=kt=knT=k1nτ+1T.

Claim: gcd(T,n)=1.
This follows from the fact that n divides all the terms of T except the first term.The first term is not divisible by n because k divides n and therefore n divides band a and b are coprime.
Claim: gcd(T,k1)=1.
This follows from the fact that k divides b, so k1 divides b, and a andb are coprime.By unique factorization of integers, then, it must be that there are positive integersq, w and λ such thatT=qn, k1=wn and nτ+1=λn. Since n is a prime, it followsthat λ=ng for some positive integer g. Hence gn=τ+1.

It follows thatbn=wnngnqn, so that b=ngwq. From a+k=b+f we get

vp+ngn-1wn=ngwq+vn

which we can regroup to get

v(p-vn-1)=ngw(q-ng(n-1)-1wn-1).

Since a and b are coprime, it follows that v and ngw are coprime.Hence

u:=p-vn-1ngw=q-ng(n-1)-1wn-1v

is an integer.It follows that

vp-vn=ngwq-ngn-1wn=nguwv

and one can now express a,b and c in terms of u,v,w.

Lemma 4. Let u be the integer of Lemma 3. There is a monic polynomialMathworldPlanetmath Pwith integer coefficientsMathworldPlanetmath such that
(a) P(u)=0,
(b) the sum of the roots of P is 0, and
(c) all coefficients of P are divisible by n except that whenβ is positive the last coefficient is not divisible by n.
Proof. We use the same cases as in Lemma 3. (1) In this case we have

(uwv+vn)n+(uwv+wn)n-(uwv+vn+wn)n=0.

The left hand side can be expanded using the binomial theorem to get apolynomialMathworldPlanetmathPlanetmath Qwith coefficients thatdepend on v and w. For the coefficient of un-i we have to combine

(ni)(uwv)n-i(vn)i+(ni)(uwv)n-i(wn)i-(ni)(uwv)n-i(vn+wn)i
=un-i(ni)(wv)n-i(-(vn+wn)i+(vn)i+(wn)i)   (5)

Clearly if i=1 this coefficient is 0.If i=0 the coefficient is (wv)n.For the other terms we can writethem as

(ni)(wv)n-ivnwnj=1i-1(ij)vn(j-1)wn(i-j-1)

so that the coefficient is divisible by (wv)n.The coefficient is also divisible by n if 1in by Lemma 1.So we set P:=Q/(wv)n to get the conclusionMathworldPlanetmath for case (1).
(2) We proceed as in case (1). The left side of the equation

(nguwv+vn)n+(nguwv+ngn-1wn)n-(nguwv+vn+ngn-1wn)n=0

can be expanded by the binomial theorem to get a polynomialQ with coefficients that depend on v and w.It is clear that the leading term is(nguwv)n. For the coefficient of un-i we have to combine

(ni)(nguwv)n-i(vn)i+(ni)(nguwv)n-i(ngn-1wn)i-(ni)(nguwv)n-i(vn+ngn-1wn)i
=un-i(ni)(ngwv)n-i((vn)i+(ngn-1wn)i-(vn+ngn-1wn)i.   (6)

This form makes it clear that the coefficient is 0 if i=1 and divisible by n if1i<n.If i=0, the coefficient is (ngwv)n. Equation (6) is equal to

un-i(ni)(ngwv)n-ij=1i-1vn(j-1)(ngn-1wn)i-j-1vnngn-1wn

which shows that ngnwnvn divides each coefficient. SetP:=Q/(ngwv)n to get the conclusion for case (2).

Lemma 5. The polynomial P of Lemma 4 has exactly one positive root.
Proof. By (5) and (6) the coefficients of P are negative except for the leadingcoefficient. So there is exactly one sign change and by Descartes’s rule of signsthere is exactly one positive root.

Definition. For each real root ui of P we can define a, b and c.(For example, a=uiwv+vn and so on.) We say that a root ui isacceptable if the resulting a,b,c are all positive integers.

Lemma 6. The only acceptable rootof P is u and u>0.
Proof. Suppose that ui is a nonpositive acceptable root. Then a,b,c are allpositive and in case (1) we have

a+b=2uiwv+vn+wn=c+uiwvc,

while in case (2) we have

a+b=2nguiwv+vn+ngn-1wn=c+nguiwvc.

But

an+bn<(a+b)ncn

which is a contradiction. Since u is acceptable, it must be thatu>0.

The following lemma 7 is incorrect.
Lemma 7. n does not divide a+b.
Proof. We use the cases of Lemma 3. (1) We write

an+bn=(a+b)Q

where

Q=j=1nan-jbj-1(-1)j-1.

It is known that the common divisorMathworldPlanetmathPlanetmath if a+b and Q is n and thatif ns||a+b then ns+1||an+bn.Hence, we can write

a+b=nsδ

and

Q=nγ

where gcd(n,δ)=1, gcd(n,γ)=1 and gcd(δ,γ)=1.From cn=an+bn=(a+b)Q=ns+1δγwe get s=n-1, n||c and

a+b=nn-1δ.   (7)

Since n divides a+b and c we have n divides 2c-(a+b)=vn+wn.It is also known that

vn+wn=(v+w)[(v+w)n-1-nvw(vn-1++wn-1)]   (8)

so that n divides v+w. But then from (8) again, n2 divides vn+wn.Now from (7) we have n2 divides a+b.Hence

2c=(a+b)+vn+wn

is divisible by n2 and this is a contradiction.
(2) In this case

a+b=2nguwv+vn+ngn-1wn

so that if n divides a+b then n divides vn and therefore n divides v.From a=vp we get then n divides a and therefore n divides b=a+b-a.But a and b are coprime, so we have a contradiction.

Since Lemma 7 is incorrect, Lemma 8 is also incorrect.
Lemma 8. There are positive integersup and cp such thata+b=upn and c=upcp.
Proof. We can write

xz+yz=(x+y)Q

where

Q=j=1zxz-jyj-1(-1)j-1.

It is an old result first attributed to Nicolas Malebranche (1638-1715)that if x and y are coprime and d is a prime divisorMathworldPlanetmathPlanetmath ofx+y and Q then d divides z. I will give a proof of this here.Define

Q1=Q=xz-1+xz-2y+j=3zxz-jyj-1(-1)j-1.

Let

Q2=Q1-xz-2(x+y)=-2xz-2y+xz-3y2+j=4zxz-jyj-1(-1)j-1.

Then d divides Q2.Define

Q3=Q2+2xz-3y(x+y)=3xz-3y2-xz-4y3+j=5zxz-jyj-1(-1)j-1.

In general

Qn=(-1)n+1nxz-nyn-1+xz-n-1yn+j=n+2zxz-jyj-1(-1)j-1

and each Qn is divisible by d. Hence,

Qz=±zyz-1

will be divisible by d. But d does not divide y (since otherwise it would also dividex, which would contradict that x and y are coprime). Hence, d divides z.Using this result, we can say thatif

an+bn=(a+b)(an-1-an-2b+-abn-2+bn-1)=(a+b)Q

then a+b and Q are coprime. Because if d is a prime divisor of eachthen d divides n, so d=n and then n divides a+b, which contradicts Lemma 7.By unique factorization there are positive integers up and cpsuch that

a+b=upn

and

Q=cpn.

Then cn=(a+b)Q=upncpn.Hence c=upcp.

Lemma 9. Let p,u,v,w be as in Lemma 3. Let cp be as inLemma 8. Suppose there are positive integers hand q such that

ah+bq=cp.

Then one of the following possibilities holds:
(a) h=hkc,q=qkc for some integers hk and qk;
(b) h=q=jcp for some integer j;
(c) h=jwn(n-1),q=jvn(n-1) for some integer j;
(d) h=jb,q=jwn for some integer j;
(e) h=jvn,q=-jwn for some integer j;
(f) h=jvn,q=j(2uwv+wn+2vn) for some integer j;
(g) h=j(2uwv+2wn+vn),q=jwn for some integer j.
Proof. At this point I think the proof is incomplete since he does not prove the result, but ratherverifies that each of the possible solutions is indeed a solution. Later on,he needs to know that theseare the only solutions.

2 n is divisible by a prime greater than 2

If n=mz where z is a prime greater than 2, then

(am)z+(bm)z=(cm)z

and we can apply the results of section 1 to conclude that no such z can exist.

3 n is a power of 2

It is known that if n=4 then Fermat’s Last Theorem is true. For example,see [1]. So if n=2t,t3, then we can write

(a2t-2)4+(b2t-2)4=(c2t-2)4

which contradicts the theorem for n=4.

References

  • 1 G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, 5th ed., OxfordUniversity Press, page 191.
随便看

 

数学辞典收录了18232条数学词条,基本涵盖了常用数学知识及数学英语单词词组的翻译及用法,是数学学习的有利工具。

 

Copyright © 2000-2023 Newdu.com.com All Rights Reserved
更新时间:2025/5/5 2:26:22