proposed elementary proof of Fermat’s last theorem
Michael Pogorsky has offered what is said to be an elementary proof of Fermat’slast theorem. 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 and such that
.
We can assume without loss of generality that and are mutually coprime, so that in factthey are also pairwise coprime. The proof is split into 3 major cases:(1) is a prime greater than 2, (2) is divisible by a prime greater than 2,and (3) is a power of 2.
1 n is a prime greater than 2
Write as
for some integers and .
Then
Using the binomial theorem we can write
and
Lemma 1. If is prime number then divides for .
The proof is easy.
Claim: .
Proof. A factor of and will also divide and by equations (2)and (3). But and are coprime, so the gcd of and must be 1.
Now write (2) as for some integer .
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 for some integer .
Lemma 2. and forsome nonnegative integers and .
Proof. Suppose divides where is a prime. Then divides and .We can write for some integer , so that divides and therefore divides . Hence divides or divides . But if divides then divides , a contradiction. Hence divides . But is a prime, so. From this we get that for some nonnegative integer .Similarly, forsome nonnegative integer .
It is clear that at least one of and is zero, otherwise divides and . Without loss of generality, we can assume that .
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 such
1) ,
2) if then
and
3) if then there is a positive integers such that
and
Proof. (1) We have ,where and are coprime. By unique factorization of integersit must be that and for some positive integers and . It follows that.
(2) Similarly, there are positive integers and such that , where .From we get
and after regrouping we have
Since it follows that , so that
and
Hence
is an integer. Using we can now write, , and .
(3) Since , we have .Since , we can write
where and . By Lemma 1 there is a positive integer such that for . We can write
where .Hence,
Claim: .
This follows from the fact that divides all the terms of except the first term.The first term is not divisible by because divides and therefore divides and and are coprime.
Claim: .
This follows from the fact that divides , so divides , and and are coprime.By unique factorization of integers, then, it must be that there are positive integers, and such that, and . Since is a prime, it followsthat for some positive integer . Hence .
It follows that, so that . From we get
which we can regroup to get
Since and are coprime, it follows that and are coprime.Hence
is an integer.It follows that
and one can now express , and in terms of ,,.
Lemma 4. Let be the integer of Lemma 3. There is a monic polynomial with integer coefficients
such that
(a) ,
(b) the sum of the roots of is 0, and
(c) all coefficients of are divisible by except that when is positive the last coefficient is not divisible by .
Proof. We use the same cases as in Lemma 3. (1) In this case we have
The left hand side can be expanded using the binomial theorem to get apolynomial with coefficients thatdepend on and . For the coefficient of we have to combine
Clearly if this coefficient is 0.If the coefficient is .For the other terms we can writethem as
so that the coefficient is divisible by .The coefficient is also divisible by if by Lemma 1.So we set to get the conclusion for case (1).
(2) We proceed as in case (1). The left side of the equation
can be expanded by the binomial theorem to get a polynomial with coefficients that depend on and .It is clear that the leading term is. For the coefficient of we have to combine
This form makes it clear that the coefficient is 0 if and divisible by if.If , the coefficient is . Equation (6) is equal to
which shows that divides each coefficient. Set to get the conclusion for case (2).
Lemma 5. The polynomial of Lemma 4 has exactly one positive root.
Proof. By (5) and (6) the coefficients of 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 of we can define , and .(For example, and so on.) We say that a root isacceptable if the resulting are all positive integers.
Lemma 6. The only acceptable rootof is and
Proof. Suppose that is a nonpositive acceptable root. Then are allpositive and in case (1) we have
while in case (2) we have
But
which is a contradiction. Since is acceptable, it must be that.
The following lemma 7 is incorrect.
Lemma 7. does not divide .
Proof. We use the cases of Lemma 3. (1) We write
where
It is known that the common divisor if and is and thatif then .Hence, we can write
and
where , and .From we get , and
Since divides and we have divides .It is also known that
so that divides . But then from (8) again, divides .Now from (7) we have divides .Hence
is divisible by and this is a contradiction.
(2) In this case
so that if divides then divides and therefore divides .From we get then divides and therefore divides .But and are coprime, so we have a contradiction.
Since Lemma 7 is incorrect, Lemma 8 is also incorrect.
Lemma 8. There are positive integers and such that and .
Proof. We can write
where
It is an old result first attributed to Nicolas Malebranche (1638-1715)that if and are coprime and is a prime divisor of and then divides . I will give a proof of this here.Define
Let
Then divides .Define
In general
and each is divisible by . Hence,
will be divisible by . But does not divide (since otherwise it would also divide, which would contradict that and are coprime). Hence, divides .Using this result, we can say thatif
then and are coprime. Because if is a prime divisor of eachthen divides , so and then divides , which contradicts Lemma 7.By unique factorization there are positive integers and such that
and
Then .Hence .
Lemma 9. Let be as in Lemma 3. Let be as inLemma 8. Suppose there are positive integers and such that
Then one of the following possibilities holds:
(a) for some integers and ;
(b) for some integer ;
(c) for some integer ;
(d) for some integer ;
(e) for some integer ;
(f) for some integer ;
(g) for some integer .
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 where is a prime greater than 2, then
and we can apply the results of section 1 to conclude that no such can exist.
3 n is a power of 2
It is known that if then Fermat’s Last Theorem is true. For example,see [1]. So if , then we can write
which contradicts the theorem for .
References
- 1 G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, 5th ed., OxfordUniversity Press, page 191.