Fermat s Great Theorem boils down to the proof of elliptic curves 30

I don't know about such questions, but if you want someone to find them for you, I suggest you increase your score to 100 or higher.

Prove Fermat's theorem (proof process explained in detail).

Known: a2+b2=c2

Let c=b+k,k=, then a 2+b 2=(b+k) 2.

Because, the integer c must be larger than both a and b, and at least greater than 1, so k=

Let a=d (n 2), b = h (n 2), c = p (n 2);

Then a 2 + b 2 = c 2 can be written as d n + h n = p n, n =

When n=1, d+h=p, d, h, and p can be arbitrary integers.

When n=2, a=d, b=h, c=p, then d2+h2=p2=> a2+b2=c2.

When n 3, a 2 = d n, b 2 = h n, c 2 = p n.

Because, a=d (n 2), b = h (n 2), c = p (n 2); In order to ensure that d, h, and p are integers, it is necessary to ensure that a, b, and c must all be perfectly squared.

a, b, and c must be squared as integers in order for d, h, and p to be integers in the formula d n + h n = p n.

If d, h, and p cannot exist as integers at the same time in the formula, then Fermat's theorem holds.

Let a=mk, then b=k(m 2-1) 2.

Let m=k, then a=m 2, b=m(m 2-1) 2, let m 2=(m 2-1), then b = (m 2) 2, c = (m 2) 2+m.

then a 2 + b 2 = c 2 => m 4 + (m 2) 4 = [(m 2) 2 + m] 2

m2(2m2-m-2)=0,m1=0(rounded),m2=(1 17)4(non-integer).

Also, when m2=(m2-1), (it is also possible to let ) b = (m2-1) 2

then a 2 + b 2 = c 2 => m 4+(m 2-1) 4 = [(m 2-1) 2+m] 2

m(m^2-1)(2m^2-m-2)=0，m1=0，m2=±1，m3=(1±√17)/4。

Validation: When m=1, b=h(n2)=(m2-1)2=0;i.e. a2=c2. It does not meet the requirements of the question.

If d, h, and p can be in the form of integers, then the equation d n+h n=p n is true, and Fermat's theorem is not true. Otherwise, d n+h n≠p n inequalities hold and Fermat's theorem holds.

Beijing, Shi Zhongxia.

This person's proof is wrong, the formula of Fermat's great theorem is an integer inequality, not an equation of irrational numbers, and it is necessary to use the general solution formula of the Pythagorean equation of integers to prove it, and Mao Guicheng has found the wonderful proof method that Fermat said.

The number to the left of the equation for the general solution of the Pythagorean integer equation is [a"-b"k, the number to the right of the equal sign is a"+b"k, these two numbers are not arrays of the same power greater than 1 power, second, the inequality formula of Fermat's great theorem is infinitely reduced to the form of 2 powers, and third, comparing the two formulas shows that Fermat's great theorem is correct.

See "Fermat's Great Theorem Cleverly Proofed". This article was published in the Journal of Shenyang Institute of Aeronautical Technology in the third issue of 2008. The full text is only four pages long, more than 3,000 words, and can be understood by anyone with a middle school level.

The article has been included in the China Journal Full-text Database, Chinese Science and Technology Journals Database, and the China Academic Journals Database--- which can be seen on the Internet. Or type "Fermat's Great Theorem Cleverly Proved" on the Internet, you can see its web page, and then register and log in to see the full text.

Hello! Fermat's theorem is that when n>2 and x*y*z≠0, x n+y n=z n has no integer solution.

It is made with the connection of the modular form and the elliptic equation.

It was solved by solving the Taniyama Shimura conjecture.

Story Approach. I hope you are satisfied with my answer! Thank you!

If you can understand that proof, you will probably be able to enter a general mathematics institute in China.

All I can say is.

This is something that elementary mathematics can't solve.

When I go to college, I can't understand it without studying mathematics.

This model is simple.

In fact, a lot of advanced mathematics is used.

It is advisable to be interested first, but do not delve into it.

x^a+y^a=z^a

When a is greater than 2, xyz has no integer solution.

A foreign strongman certificate came out.

When n is greater than or equal to 3

an+bn=cn is not true.

It was Wells who proved it.

The proof given by the owner was originally intended to use the method of counterproof.

First of all, we must know that in order to prove a proposition by the method of counterproof, we should first assume that its negative proposition is true, and then derive a contradiction through deduction, so as to know that the hypothetical negative proposition is not true, that is, the original proposition is true.

To describe the original proposition given in this question in precise mathematical language, it should be:

For any non-zero integer a, b, c, and any natural number n greater than 3, a n + b n is not equal to c n

The corresponding negative proposition should be:

There is a definite set of non-zero integers a, b, c, and a natural number n greater than 3, satisfying a n + b n = c n

It should be noted"Optional"The negation is"exists", ie"P is true in any case"The negative proposition is"There is a circumstance that makes p untrue")

Using the method of counterproof, it is assumed that the negative proposition is true.

Square the two sides of the equation to get a (2n)+2*a n*b n+b (2n)=c (2n), and the proof of the Lord is right so far.

And in the next step 3, "because 2n belongs to n, the two sides of the equation subtract the 2n power of a, the 2n power of b and the 2n power of c, and we get: twice the nth power of a multiplied by the nth power of b equals zero", this sentence is wrong and unfounded. Because in the assumption at the beginning of the counter-argument, we are saying that the equations are true for a set of definite a, b, c, and n, and not for any n-equation, so we cannot get a (2n) + b (2n) = c (2n), so we cannot eliminate the correlation on both sides of the equation to get an equation such as 2*a n*b n=0.

This is where the error lies in the proof given by the Lord. The key is that the original proposition and its corresponding negative proposition are not clearly seen when using the counterargument method, which leads to the wrong application of assumptions and makes the proof wrong.