# Interesting lemma

Interesting lemma

If P.Ferma knew the proof of the following lemma
"Lemma. If numbers a, b, c, have no common factor, numbers c-a and c-b are also mutually-prime and n is odd, then the numbers (c^n-a^n)/(c-a) and (c^n-b^n)/(c-b) are also mutually-prime",
then with its aid it is possible to briefly and simply prove Fermat's last theorem.

Actually, in the Fermat’s equality (where numbers a, b, c have no common factor and n is odd) numbers c-a and c-b, obviously, mutually-prime. And then from the Fermat's little theorem it follows that with prime q>2c the numbers c^(q-1)-a^(q-1) and c^(q-1)-b^(q-1) are multiple by q.
And since, according to lemma, the numbers (c^(q-1)-a^(q-1))/(c-a) and (c^(q-1)-b^(q-1))/(c-b) are mutually-prime (i.e. have no common factor), then one of the numbers c-a and c-b is divided by q (>2c>c-b>c-a), i.e., the solution of the Fermat’s equation is not integer.

It remains to learn, who and when proved lemma.

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Hurkyl
Staff Emeritus
Gold Member
I hope nobody proved that lemma; it's not true!

Here's the smallest counterexample I could find:

a = 2
b = 2
c = 3
n = 3

a, b, c don't have a common factor.

c-a = 1
c-b = 1

These don't have a common factor

(c^n - a^n) / (c-a) = 19
(c^n - b^n) / (c-b) = 19

These do have a common factor.

Here's the smallest I could find with a and b unequal:

a = 1
b = 4
c = 9
n = 3

These don't have a common factor.

c-a = 8
c-b = 5

These don't have a common factor.

(c^3 - a^3) / (c-a) = 91
(c^3 - b^3) / (c-b) = 133

These do have a common factor: 7.

Here's another:

a = 5
b = 6
c = 17
n = 3

c-a = 12
c-b = 11

(c^3 - a^3) / (c-a) = 399
(c^3 - b^3) / (c-b) = 427

These do have a common factor: 7.

I hope nobody proved that lemma; it's not true!

Here's the smallest I could find with a and b unequal:

a = 1
b = 4
c = 9
n = 3

These don't have a common factor.

c-a = 8
c-b = 5

These don't have a common factor.

(c^3 - a^3) / (c-a) = 91
(c^3 - b^3) / (c-b) = 133

These do have a common factor: 7.
Outstanding counterexamples!

Therefore I make correction to the Lemma:
By the way, now to prove the lemma it is much easier.

matt grime
Homework Helper
Then feel free to post the proof.

Lemma

Then feel free to post the proof.
Here are the minimum requirements for the lemma, necessary for the brief proof of the FLT: "Lemma. If integers a, b, c have only one common divisor 1, a+b=c and numbers (c^n-a^n) and (c^n-b^n) have common divisor d>2, then with n>1 one of the numbers c-b, c-a, a+b, c+b, c+a, a-b is divided by d ".

Hurkyl
Staff Emeritus
Gold Member
Outstanding counterexamples!

Therefore I make correction to the Lemma:
By the way, now to prove the lemma it is much easier.
No it's not.

a = 3, b = 10, c = 13, n = 3 is another counterexample.
a = 5, b = 12, c = 17, n = 3 is another.

If you're guessing, call it a conjecture. You can't call something a lemma until you have actually proven it.

Here are the minimum requirements for the lemma, necessary for the brief proof of the FLT: ...
matt asked you to post the proof. That's not a proof.

No it's not.

a = 3, b = 10, c = 13, n = 3 is another counterexample.
a = 5, b = 12, c = 17, n = 3 is another.
10-3=7,
12-5=7.
Consequently, it is possible to begin the search for its proof.

10-3=7,
12-5=7.
Consequently, it is possible to begin the search for its proof.
What do you mean by accuracy of the lemma? If there exists A counter example then it isn't true.

What do you mean by accuracy of the lemma? If there exists A counter example then it isn't true.
Here is the Lemma, which easily proves and which is sufficient for the brief proof of the FLT:
Lemma. If integers a, b, c have only one common divisor 1, a+b=c, number (c^n-a^n) and (c^n-b^n) have common prime divisor d>3c^2, in base d numbers a^n, b^n, c^n finish by digit 1, then with n>1 either number (a^n-b^n)/(a+b) is not divided by d, or number 3c^2 is divided into d ".

to be continued

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Lemma And Its Proof

to be continued
Lemma.
If integers a, b, c have only one common divisor 1, a+b=c, and number (c^n-a^n) and (c^n-b^n) have common prime divisor d>3c^2, and in base d numbers a^n, b^n, c^n finish by digit 1, then with n>1 either number (a^n-b^n)/(a+b) is not divided by d, or number 3c^2 is divided by d.

Proof.
Let us assume that number (a^n-b^n)/(a+b) is divided by d.
Then number S=(c^n-b^n)/(c-b)+(c^n-a^n)/(c-a)+(a^n-b^n)/(a+b), or
(c^n-b^n)/a+(c^n-a^n)/b+(a^n-b^n)/c = [c^(n+2)+b^(n+2)+(a+b)c^(n+1]/abc=
=[c^2+b^2+a(c+b)]/abc, is divided by d.
But since number a^n, b^n, c^n finish by number 1, then the number s = c^2+b^2+a(c+b), where 0<s<d, is single-digit number and, therefore, number S is NOT DIVIDED by d.

Elementary proof of Fermat's last theorem

Lemma.
If integers a, b, c have only one common divisor 1, a+b=c, and number (c^n-a^n) and (c^n-b^n) have common prime divisor d>3c^2, and in base d numbers a^n, b^n, c^n finish by digit 1, then with n>1 either number (a^n-b^n)/(a+b) is not divided by d, or number 3c^2 is divided by d.
Elementary proof of Fermat's last theorem

According to known theorem from the theory of integers, for any prime n there exist infinite great number of the prime numbers d of form d=pn+1, where p is integer. (This theorem is the simple consequence from the fact that, if numbers a and b are mutually-prime and n prime, then each divisor of the number (a^n+b^n)/(a+b) takes form pn+1).

Proof FLT:
Let us take simple d>c^(4n). Then, according to Fermat's little theorem, the numbers C^p-B^p, C^p-A^p, A^p-B^p, where C=c^n, B=b^n; A=a^n, are divided by d. And since A+B=C, then, according to Lemma, number 3C^2 is divided by d>3C^2. Consequently, the solution of Fermat’s equation is not integer.

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arildno
Homework Helper
Gold Member
Dearly Missed
Can't you just stop posting this nonsense?

Can't you just stop posting this nonsense?
OK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Lemma is not true

Lemma.
If integers a, b, c have only one common divisor 1, a+b=c, and number (c^n-a^n) and (c^n-b^n) have common prime divisor d>3c^2, and in base d numbers a^n, b^n, c^n finish by digit 1, then with n>1 either number (a^n-b^n)/(a+b) is not divided by d, or number 3c^2 is divided by d.
Lemma is not true. New version of the lemma:
If in the simple base q the mutually-simple numbers A^p, B^p, C^p (where C=c^n, B=b^n, A=a^n, prime n>2 and p=/q-1) finish by digit 1, then either A+B and A-B or C+B and C-B, or C+A and C-A have common divisor not equal to 1.
Today I have not its proof.

Time out

Hurkyl
Staff Emeritus