Please help , need to prove that x^n + y^n = z^n has no non-zero integer solutions for x, y and z when n > 2

Answer 1

I have a discovered a truly marvelous proof of this proposition which this post is too narrow to contain.

Answer 2

No situation Ben, enable m be an integer so as that we could write x=am+b, y=cm+d and z=em+f. we are going to could desire to % m>0 and under the smallest of x, y, or z so as that m> b^n +d^n-f^n which sounds like a complicated technique so i will go away that to you. Now replace the values into x^n+y^n=z^n, which we assume to be genuine and we get (am+b)^n + (cm+d)^n = (em+f)^n and if we cut back this mod m we get, b^n +d^n = f^n mod m that's b^n + d^n - f^n = 0 mod m. considering m> b^n + d^n - f^n we've not have been given any decision yet to assign, b^n + d^n - f^n = 0 or purely b^n + d^n = f^n. Please be conscious that b, d, and f are smaller than x, y and z respectively, and we are going to word the same technique returned to the numbers b, d,and f to get yet another equation with smaller values say g,i,and ok so as that g^n + i^n = ok^n and we are going to (you are able to) do this invariably yet we actually would be unable to because of the fact x,y,and z have been finite integers and so we've a contradiction. word that in spite of if the multiple x,y,or z have been unfavourable, shall we rearrange them to have an equation with effective powers and word our approach. i've got skipped over some information however the top is purely what you required. I heard that this situation grow to be known yet i do no longer understand what all of the fuss grow to be approximately. humorousness everybody?

Answer 3

No problem Ben, let m be an integer so that we may write
x=am+b, y=cm+d and z=em+f. We'll need to choose m>0 and less than the smallest of x, y, or z
so that m> b^n +d^n-f^n which looks like a complicated procedure so i'll leave that to you. Now substitute the values
into x^n+y^n=z^n, which we assume to be true and we get

(am+b)^n + (cm+d)^n = (em+f)^n and if we reduce this

mod m we get, b^n +d^n = f^n mod m which is

b^n + d^n - f^n = 0 mod m.

Since m> b^n + d^n - f^n we have no choice but to
assign,

b^n + d^n - f^n = 0 or just

b^n + d^n = f^n.

Please notice that b, d, and f are smaller than x, y and z
respectively, and we'll apply the same procedure again
to the numbers b, d,and f to get another equation with smaller values say g,i,and k so that

g^n + i^n = k^n

and we'll (you can) do this forever but we really can't because x,y,and z were finite integers and so we have a contradiction. Note that even if some of the x,y,or z were negative, we could rearrange them to have an equation
with positive powers and apply our method. I've left out some details but the conclusion is just what you required.

I heard that this problem was famous but i don't know what all the fuss was about.

Sense of humor anyone?

Answer 4

What you are asking to prove is known as Fermat's last theorem, one of the greatest problems in the history of mathematics.

It is a theorem that was made by the mathematician Fermat in 1637, but it was not until 1994 that it was finally proven. The actual proof involves very complex mathematics, well beyond what can be entered into this little text box. If you are interested, there is more information in the following link:
http://en.wikipedia.org/wiki/Fermat's_last_theorem

Answer 5

You're a funny guy, Ben. Great comeback by Pascal.