Michael Pogorsky has offered what is said to be an elementary proof of Fermat’s last theorem. Is the proof correct? The intent of 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.
proof We assume that there are positive integers and such that
We can assume without loss of generality that and are mutually coprime, so that in fact they 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
for some integers and .
Using the binomial theorem we can write
Lemma 1. If is prime number then divides for .
The proof is easy.
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 unclear so 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 for
some 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, for some 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 these
Case 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
2) if then
3) if then there is a positive integers such that
Proof. (1) We have ,where and are coprime. By unique factorization of integers
it 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
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,
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.
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 follows that 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
(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 a polynomial with coefficients that depend 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 write them 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 leading coefficient. So there is exactly one sign change and by Descartes’s rule of signs there 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 is acceptable if the resulting are all positive integers.
Lemma 6. The only acceptable root
of is and
Proof. Suppose that is a nonpositive acceptable root. Then are all positive and in case (1) we have
while in case (2) we have
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
It is known that the common divisor of and is and that if then . Hence, we can write
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
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
Then divides . Define
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 that if
then and are coprime. Because if is a prime divisor of each then divides , so and then divides , which contradicts Lemma 7. By unique factorization there are positive integers and such that
Then . Hence .
Lemma 9. Let be as in Lemma 3. Let be as in Lemma 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 rather verifies that each of the possible solutions is indeed a solution. Later on,he needs to know that these are 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 . So if , then we can write
which contradicts the theorem for .
- 1 G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford University Press, page 191.