proposed elementary proof of Fermat’s last theorem

proposed elementary proof of Fermat’s last theorem

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

a

,

b

and

c

such that

a

n

+

b

n

=

c

n

(

1

)

.

We can assume without loss of generality that

a

,

b

and

c

are

mutually coprime

, so that in fact
they 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

a

n

+

b

n

=

(

a

+

k

)

n

=

(

b

+

f

)

n

.

Using the

binomial theorem

we can write

a

n

=

f

⁢

(

n

⁢

b

n

-

1

+

1

2

⁢

n

⁢

(

n

-

1

)

⁢

b

n

-

2

⁢

f

+

⋯

+

f

n

-

1

)

(

2

)

and

b

n

=

k

(

n

a

n

-

1

+

1

2

n

(

n

-

1

)

a

n

-

2

k

+

⋯

+

k

n

-

1

)

.

(

3

)

Lemma 1. If

n

is

prime number

then

n

divides

(

n

k

)

for

0

<

k

<

n

.

The proof is easy.

Claim:

g

⁢

c

⁢

d

⁢

(

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

coprime

, so the gcd of

f

and

k

must be 1.

Now write (2) as

a

n

=

f

⁢

s

for some integer

s

.

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

b

n

=

k

⁢

t

for some integer

t

.

Lemma 2.

g

⁢

c

⁢

d

⁢

(

f

,

s

)

=

n

α

and

g

⁢

c

⁢

d

⁢

(

k

,

t

)

=

n

β

for
some

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

=

n

⁢

b

n

-

1

+

f

⁢

T

for some integer

T

, so that

q

divides

s

-

f

⁢

T

and therefore

q

divides

n

⁢

b

n

-

1

. Hence

q

divides

n

or

q

divides

b

n

-

1

. But if

q

divides

b

n

-

1

then

q

divides

b

, a

contradiction

. Hence

q

divides

n

. But

n

is a prime, so

q

=

n

. From this we get that

g

⁢

c

⁢

d

⁢

(

f

,

s

)

=

n

α

for some nonnegative integer

α

.
Similarly,

g

⁢

c

⁢

d

⁢

(

k

,

t

)

=

n

β

for
some nonnegative integer

β

.

It is clear that at least one of

α

and

β

is zero, otherwise

n

divides

f

and

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 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

p

,

u

,

v

,

w

such

1)

a

=

v

⁢

p

,

2) if

β

=

0

then

a

=

u

⁢

w

⁢

v

+

v

n

b

=

u

⁢

w

⁢

v

+

w

n

c

=

u

⁢

w

⁢

v

+

v

n

+

w

n

.

and

3) if

β

>

0

then there is a positive integers

g

such that

a

=

n

g

⁢

u

⁢

w

⁢

v

+

v

n

b

=

n

g

⁢

u

⁢

w

⁢

v

+

n

g

⁢

n

-

1

⁢

w

n

and

c

=

n

g

⁢

u

⁢

w

⁢

v

+

v

n

+

n

g

⁢

n

-

1

⁢

w

n

.

Proof. (1) We have

a

n

=

f

⁢

s

,where

f

and

s

are coprime. By unique factorization of integers
it must be that

f

=

v

n

and

s

=

p

n

for some positive integers

p

and

v

. It follows that

a

=

p

⁢

v

.

(2) Similarly, there are positive integers

w

and

q

such that

b

=

w

⁢

q

, where

w

n

=

k

.
From

a

+

k

=

b

+

f

we get

v

⁢

p

+

w

n

=

w

⁢

q

+

v

n

and after regrouping we have

v

⁢

(

p

-

v

n

-

1

)

=

w

⁢

(

q

-

w

n

-

1

)

Since

g

⁢

c

⁢

d

⁢

(

f

,

k

)

=

1

it follows that

g

⁢

c

⁢

d

⁢

(

v

,

w

)

=

1

, so that

v

|

(

q

-

w

n

-

1

)

and

w

|

(

p

-

v

n

-

1

)

.

Hence

u

:=

p

-

v

n

-

1

w

=

q

-

w

n

-

1

v

(

4

)

is an integer. Using

u

we can now write

a

=

u

⁢

w

⁢

v

+

v

n

,

b

=

u

⁢

w

⁢

v

+

w

n

, and

c

=

u

⁢

w

⁢

v

+

v

n

+

w

n

.

(3) Since

α

=

0

, we have

g

⁢

c

⁢

d

⁢

(

f

,

s

)

=

1

.
Since

β

>

0

, we can write

k

=

k

1

⁢

n

τ

where

τ

>

0

and

g

⁢

c

⁢

d

⁢

(

k

1

,

n

)

=

1

. By Lemma 1 there is a positive integer

c

i

such that

(

n

i

)

=

n

⁢

c

i

for

0

<

i

<

n

. We can write

t

=

∑

i

=

1

n

(

n

i

)

⁢

a

n

-

i

⁢

k

i

-

1

=

∑

i

=

1

n

-

1

n

⁢

c

i

⁢

a

n

-

i

⁢

k

i

-

1

+

k

n

-

1

=

∑

i

=

1

n

-

1

n

⁢

c

i

⁢

a

n

-

i

⁢

k

i

-

1

+

n

⁢

k

1

⁢

n

τ

-

1

⁢

k

n

-

2

=

n

⁢

T

where

T

=

a

n

-

1

+

1

2

⁢

(

n

-

1

)

⁢

a

n

-

2

⁢

k

+

⋯

+

k

1

⁢

n

τ

-

1

⁢

k

n

-

2

.
Hence,

b

n

=

k

⁢

t

=

k

⁢

n

⁢

T

=

k

1

⁢

n

τ

+

1

⁢

T

.

Claim:

g

⁢

c

⁢

d

⁢

(

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

b

and

a

and

b

are coprime.

Claim:

g

⁢

c

⁢

d

⁢

(

T

,

k

1

)

=

1

.

This follows from the fact that

k

divides

b

, so

k

1

divides

b

, and

a

and

b

are coprime.
By unique factorization of integers, then, it must be that there are positive integers

q

,

w

and

λ

such that

T

=

q

n

,

k

1

=

w

n

and

n

τ

+

1

=

λ

n

. Since

n

is a prime, it follows
that

λ

=

n

g

for some positive integer

g

. Hence

g

⁢

n

=

τ

+

1

.

It follows that

b

n

=

w

n

⁢

n

g

⁢

n

⁢

q

n

, so that

b

=

n

g

⁢

w

⁢

q

. From

a

+

k

=

b

+

f

we get

v

⁢

p

+

n

g

⁢

n

-

1

⁢

w

n

=

n

g

⁢

w

⁢

q

+

v

n

which we can regroup to get

v

⁢

(

p

-

v

n

-

1

)

=

n

g

⁢

w

⁢

(

q

-

n

g

⁢

(

n

-

1

)

-

1

⁢

w

n

-

1

)

.

Since

a

and

b

are coprime, it follows that

v

and

n

g

⁢

w

are coprime.
Hence

u

:=

p

-

v

n

-

1

n

g

⁢

w

=

q

-

n

g

⁢

(

n

-

1

)

-

1

⁢

w

n

-

1

v

is an integer.
It follows that

v

⁢

p

-

v

n

=

n

g

⁢

w

⁢

q

-

n

g

⁢

n

-

1

⁢

w

n

=

n

g

⁢

u

⁢

w

⁢

v

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 polynomial

P

with integer

coefficients

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 when

β

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

(

u

⁢

w

⁢

v

+

v

n

)

n

+

(

u

⁢

w

⁢

v

+

w

n

)

n

-

(

u

⁢

w

⁢

v

+

v

n

+

w

n

)

n

=

0

.

The left hand side can be expanded using the binomial theorem to get a

polynomial

Q

with coefficients that
depend on

v

and

w

. For the coefficient of

u

n

-

i

we have to combine

(

n

i

)

⁢

(

u

⁢

w

⁢

v

)

n

-

i

⁢

(

v

n

)

i

+

(

n

i

)

⁢

(

u

⁢

w

⁢

v

)

n

-

i

⁢

(

w

n

)

i

-

(

n

i

)

⁢

(

u

⁢

w

⁢

v

)

n

-

i

⁢

(

v

n

+

w

n

)

i

=

u

n

-

i

⁢

(

n

i

)

⁢

(

w

⁢

v

)

n

-

i

⁢

(

-

(

v

n

+

w

n

)

i

+

(

v

n

)

i

+

(

w

n

)

i

)

(

5

)

Clearly if

i

=

1

this coefficient is 0.
If

i

=

0

the coefficient is

(

w

⁢

v

)

n

.
For the other terms we can write
them as

(

n

i

)

⁢

(

w

⁢

v

)

n

-

i

⁢

v

n

⁢

w

n

⁢

∑

j

=

1

i

-

1

(

i

j

)

⁢

v

n

⁢

(

j

-

1

)

⁢

w

n

⁢

(

i

-

j

-

1

)

so that the coefficient is divisible by

(

w

⁢

v

)

n

.
The coefficient is also divisible by

n

if

1

≤

i

≤

n

by Lemma 1.
So we set

P

:=

Q

/

(

w

⁢

v

)

n

to get the

conclusion

for case (1).

(2) We proceed as in case (1). The left side of the equation

(

n

g

⁢

u

⁢

w

⁢

v

+

v

n

)

n

+

(

n

g

⁢

u

⁢

w

⁢

v

+

n

g

⁢

n

-

1

⁢

w

n

)

n

-

(

n

g

⁢

u

⁢

w

⁢

v

+

v

n

+

n

g

⁢

n

-

1

⁢

w

n

)

n

=

0

can be expanded by the binomial theorem to get a polynomial

Q

with coefficients that depend on

v

and

w

.
It is clear that the leading term is

(

n

g

⁢

u

⁢

w

⁢

v

)

n

. For the coefficient of

u

n

-

i

we have to combine

(

n

i

)

⁢

(

n

g

⁢

u

⁢

w

⁢

v

)

n

-

i

⁢

(

v

n

)

i

+

(

n

i

)

⁢

(

n

g

⁢

u

⁢

w

⁢

v

)

n

-

i

⁢

(

n

g

⁢

n

-

1

⁢

w

n

)

i

-

(

n

i

)

⁢

(

n

g

⁢

u

⁢

w

⁢

v

)

n

-

i

⁢

(

v

n

+

n

g

⁢

n

-

1

⁢

w

n

)

i

=

u

n

-

i

(

n

i

)

(

n

g

w

v

)

n

-

i

(

(

v

n

)

i

+

(

n

g

⁢

n

-

1

w

n

)

i

-

(

v

n

+

n

g

⁢

n

-

1

w

n

)

i

.

(

6

)

This form makes it clear that the coefficient is 0 if

i

=

1

and divisible by

n

if

1

≤

i

<

n

.
If

i

=

0

, the coefficient is

(

n

g

⁢

w

⁢

v

)

n

. Equation (6) is equal to

u

n

-

i

⁢

(

n

i

)

⁢

(

n

g

⁢

w

⁢

v

)

n

-

i

⁢

∑

j

=

1

i

-

1

v

n

⁢

(

j

-

1

)

⁢

(

n

g

⁢

n

-

1

⁢

w

n

)

i

-

j

-

1

⁢

v

n

⁢

n

g

⁢

n

-

1

⁢

w

n

which shows that

n

g

⁢

n

⁢

w

n

⁢

v

n

divides each coefficient. Set

P

:=

Q

/

(

n

g

⁢

w

⁢

v

)

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 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

u

i

of

P

we can define

a

,

b

and

c

.
(For example,

a

=

u

i

⁢

w

⁢

v

+

v

n

and so on.) We say that a root

u

i

is

acceptable

if the resulting

a

,

b

,

c

are all positive integers.

Lemma 6. The only acceptable root
of

P

is

u

and

u

>

0

.

Proof. Suppose that

u

i

is a nonpositive acceptable root. Then

a

,

b

,

c

are all
positive and in case (1) we have

a

+

b

=

2

⁢

u

i

⁢

w

⁢

v

+

v

n

+

w

n

=

c

+

u

i

⁢

w

⁢

v

≤

c

,

while in case (2) we have

a

+

b

=

2

⁢

n

g

⁢

u

i

⁢

w

⁢

v

+

v

n

+

n

g

⁢

n

-

1

⁢

w

n

=

c

+

n

g

⁢

u

i

⁢

w

⁢

v

≤

c

.

But

a

n

+

b

n

<

(

a

+

b

)

n

≤

c

n

which is a contradiction. Since

u

is acceptable, it must be that

u

>

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

a

n

+

b

n

=

(

a

+

b

)

⁢

Q

where

Q

=

∑

j

=

1

n

a

n

-

j

⁢

b

j

-

1

⁢

(

-

1

)

j

-

1

.

It is known that the common

divisor

if

a

+

b

and

Q

is

n

and that
if

n

s

|

|

a

+

b

then

n

s

+

1

|

|

a

n

+

b

n

.
Hence, we can write

a

+

b

=

n

s

⁢

δ

and

Q

=

n

⁢

γ

where

g

⁢

c

⁢

d

⁢

(

n

,

δ

)

=

1

,

g

⁢

c

⁢

d

⁢

(

n

,

γ

)

=

1

and

g

⁢

c

⁢

d

⁢

(

δ

,

γ

)

=

1

.
From

c

n

=

a

n

+

b

n

=

(

a

+

b

)

⁢

Q

=

n

s

+

1

⁢

δ

⁢

γ

we get

s

=

n

-

1

,

n

|

|

c

and

a

+

b

=

n

n

-

1

δ

.

(

7

)

Since

n

divides

a

+

b

and

c

we have

n

divides

2

⁢

c

-

(

a

+

b

)

=

v

n

+

w

n

.
It is also known that

v

n

+

w

n

=

(

v

+

w

)

⁢

[

(

v

+

w

)

n

-

1

-

n

⁢

v

⁢

w

⁢

(

v

n

-

1

+

⋯

+

w

n

-

1

)

]

(

8

)

so that

n

divides

v

+

w

. But then from (8) again,

n

2

divides

v

n

+

w

n

.
Now from (7) we have

n

2

divides

a

+

b

.
Hence

2

⁢

c

=

(

a

+

b

)

+

v

n

+

w

n

is divisible by

n

2

and this is a contradiction.

(2) In this case

a

+

b

=

2

⁢

n

g

⁢

u

⁢

w

⁢

v

+

v

n

+

n

g

⁢

n

-

1

⁢

w

n

so that if

n

divides

a

+

b

then

n

divides

v

n

and therefore

n

divides

v

.
From

a

=

v

⁢

p

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 integers

u

p

and

c

p

such that

a

+

b

=

u

p

n

and

c

=

u

p

⁢

c

p

.

Proof. We can write

x

z

+

y

z

=

(

x

+

y

)

⁢

Q

where

Q

=

∑

j

=

1

z

x

z

-

j

⁢

y

j

-

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 divisor

of

x

+

y

and

Q

then

d

divides

z

. I will give a proof of this here.
Define

Q

1

=

Q

=

x

z

-

1

+

x

z

-

2

⁢

y

+

∑

j

=

3

z

x

z

-

j

⁢

y

j

-

1

⁢

(

-

1

)

j

-

1

.

Let

Q

2

=

Q

1

-

x

z

-

2

⁢

(

x

+

y

)

=

-

2

⁢

x

z

-

2

⁢

y

+

x

z

-

3

⁢

y

2

+

∑

j

=

4

z

x

z

-

j

⁢

y

j

-

1

⁢

(

-

1

)

j

-

1

.

Then

d

divides

Q

2

.
Define

Q

3

=

Q

2

+

2

⁢

x

z

-

3

⁢

y

⁢

(

x

+

y

)

=

3

⁢

x

z

-

3

⁢

y

2

-

x

z

-

4

⁢

y

3

+

∑

j

=

5

z

x

z

-

j

⁢

y

j

-

1

⁢

(

-

1

)

j

-

1

.

In general

Q

n

=

(

-

1

)

n

+

1

⁢

n

⁢

x

z

-

n

⁢

y

n

-

1

+

x

z

-

n

-

1

⁢

y

n

+

∑

j

=

n

+

2

z

x

z

-

j

⁢

y

j

-

1

⁢

(

-

1

)

j

-

1

and each

Q

n

is divisible by

d

. Hence,

Q

z

=

±

z

⁢

y

z

-

1

will be divisible by

d

. But

d

does not divide

y

(since otherwise it would also divide

x

, which would contradict that

x

and

y

are coprime). Hence,

d

divides

z

.
Using this result, we can say that
if

a

n

+

b

n

=

(

a

+

b

)

⁢

(

a

n

-

1

-

a

n

-

2

⁢

b

+

⋯

-

a

⁢

b

n

-

2

+

b

n

-

1

)

=

(

a

+

b

)

⁢

Q

then

a

+

b

and

Q

are coprime. Because if

d

is a prime divisor of each
then

d

divides

n

, so

d

=

n

and then

n

divides

a

+

b

, which contradicts Lemma 7.
By unique factorization there are positive integers

u

p

and

c

p

such that

a

+

b

=

u

p

n

and

Q

=

c

p

n

.

Then

c

n

=

(

a

+

b

)

⁢

Q

=

u

p

n

⁢

c

p

n

.
Hence

c

=

u

p

⁢

c

p

.

Lemma 9. Let

p

,

u

,

v

,

w

be as in Lemma 3. Let

c

p

be as in
Lemma 8. Suppose there are positive integers

h

and

q

such that

a

⁢

h

+

b

⁢

q

=

c

⁢

p

.

Then one of the following possibilities holds:

(a)

h

=

h

k

⁢

c

,

q

=

q

k

⁢

c

for some integers

h

k

and

q

k

;

(b)

h

=

q

=

j

⁢

c

p

for some integer

j

;

(c)

h

=

j

⁢

w

n

⁢

(

n

-

1

)

,

q

=

j

⁢

v

n

⁢

(

n

-

1

)

for some integer

j

;

(d)

h

=

j

⁢

b

,

q

=

j

⁢

w

n

for some integer

j

;

(e)

h

=

j

⁢

v

n

,

q

=

-

j

⁢

w

n

for some integer

j

;

(f)

h

=

j

⁢

v

n

,

q

=

j

⁢

(

2

⁢

u

⁢

w

⁢

v

+

w

n

+

2

⁢

v

n

)

for some integer

j

;

(g)

h

=

j

⁢

(

2

⁢

u

⁢

w

⁢

v

+

2

⁢

w

n

+

v

n

)

,

q

=

j

⁢

w

n

for some integer

j

.

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

n

=

m

⁢

z

where

z

is a prime greater than 2, then

(

a

m

)

z

+

(

b

m

)

z

=

(

c

m

)

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

=

2

t

,

t

≥

3

, then we can write

(

a

2

t

-

2

)

4

+

(

b

2

t

-

2

)

4

=

(

c

2

t

-

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., Oxford
University Press, page 191.

Title

proposed elementary proof of Fermat’s last theorem

Canonical

name

ProposedElementaryProofOfFermatsLastTheorem

Date of creation

2013-03-22 17:36:55

Last modified on

2013-03-22 17:36:55

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

78

Author

Mathprof (13753)

Entry type

Proof

Classification

msc 11D41

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