Got a quick question. Show that if the integers x,y,z for a Pythagorean Triple, and if n is an integer greater than 2, then x^n+y^n != z^n. I know its a form of Fermat's Last Theorem, but I do not know how to do it.
The proof is too complicated. It used elliptic curves, modular forms and algebraic geometry.
However, I believe that one day, after much work had been done in modular forms. The proof will be short (but still hard). Let me tell you why... (a histrory lesson).
The Jordan-Closed Curve theorem, is extremely long in its original proof. But now it is a paragraph long (I believe).
The Fundamental Theorem of Algebra is extremely long (27 pages). In its original proof, by Gauss. But with the use of complex analysis the proof is not even 3 sentences.
Quadradic reciprocity first shown by Gauss (and imperfectly by Legendre) was also long and ugly. But later in his life Gauss offered a 7th and final prove of this theorem, which is about 3 pages long.
Math histroy shows us that if you wait at most 100 years, the proof will be probably a paragraph.
Not even. It is a specilized topic in mathematics. I once looked at the math department for Harvard, just to see what they have, and the closest course they offer is "Introduction to Analytic Number Theory". This topic is even beyond that.Originally Posted by putnamwinner
But hopefully that is something I will know one day.
Suppose (x,y,z) is a Pythagorean triple (that is x,y and z are in Z+, and x^2+y^2=z^2), and n>2. Then:
z^n=(z^2)^(n/2)=(x^2+y^2)^(n/2),
but n/2>1, so:
z^n=(z^2)^(n/2)=(x^2+y^2)^(n/2)>=x^n+y^n,
with equality holding only if x or y =0, which is not the case so:
z^n>x^n+y^n.
Which proves that x^n+y^n != z^n wnen (x,y,z) are a Pythagorean triple.
RonL