Recent content by mente oscura

Hello. Regards.

Hello. Regards.

Hello. Regards.

Hello. Regards.

Hello. I know that the idea is not very brilliant, but: Regards.

Hello. Yes. It is that I am a bit lazy. Regards.

Hello. Regards.

I sit it. I thought that it was a doubt of someone, but, later, I saw that it was a game. Because of it I edited, to add the spoiler. Regards.

Hello. Regards.

Hello. I have considered that "g" or "f" is an even number. Only one. You can choose, what is the pair, and see how the result is similar. In this case, the show is trivial: Let \ i \ , \ j \in{\mathbb{N}} \ / one \ is \ even \ and \ the \ other \ odd If \ k |(i+j) \ and \ k |(i-j)...

Hello. where is the problem? b=2mn=2uv=2xyzpqr According to your example: m=pqr n=xyz u=pxy v=qrz We operate: c=m^2+n^2=p^2q^2r^2+x^2y^2z^2 c=u^2-v^2=p^2x^2y^2-q^2r^2z^2 p^2q^2r^2+x^2y^2z^2=p^2x^2y^2-q^2r^2z^2 q^2r^2(p^2+z^2)=x^2y^2(p^2-z^2) q^2r^2=p^2-z^2 x^2y^2=p^2+z^2 Two...

Hello. I share with you , a demonstration, which authorship is mine. I do not know, if it exists, similar other one.Section A) demonstration Fermat’s Last Theorem, for n = 4. Let \ A, \ B, \ C \ \in{\mathbb{N}} / A, \ C \ = \ odd; \ B \ = \ even \ / A^2+B^2=C^2 (*)We consider, as possible...

Hello, Idahl. Thank you, for taking part in the challenge. But, what calculations have you realized? (Muscle) ? Regards. (Med venlig hilsen) :rolleyes:

Hello.:) Find the 6 reals roots: P(x)=x^6-25x^5-192x^4+7394x^3-48936x^2+113304x-79488 Regards.

Hello. My solution: Regards.