abc conjecture proved? (1 Viewer)

[lolcakes52]

Pretty straight forward for n=3. Just factorise x^3 + y^3 + z^3... I think. We did it on tuesday in four unit.

 

[Carrotsticks]

Pretty straight forward for n=3. Just factorise x^3 + y^3 + z^3... I think. We did it on tuesday in four unit.

I think you meant x^2 + y^2 + z^2 - (xy + yz + xz)

 

[barbernator]

I think you meant x^2 + y^2 + z^2 - (xy + yz + xz)

i think he means 2(x^3+y^3+z^3) = > x^2y + y^2x +.....+ z^2x = > xyz(......) = > 6xyz

then let x = a^1/3 and bingo.

 

[soloooooo]

An impressive feat if the papers hold up to scrutiny.

 

[Carrotsticks]

i think he means 2(x^3+y^3+z^3) = > x^2y + y^2x +.....+ z^2x = > xyz(......) = > 6xyz

then let x = a^1/3 and bingo.

Oh I had a slightly different proof in mind.