at 9:00 (and similar following steps) What if let's say x1 = 2 x0 then x1-x0 = x0 and the difference has blue color. But there is no triplet because the three numbers are in fact only two numbers. With the same setup, the last remaining color could have two numbers as well... unless somehow this constellation cannot occur?
Hi @deinauge7894, That's an excellent question, and I missed this subtlety. But going over the proof again, I see that the numbers x,y, and z in Schur's lemma need not be distinct. In other words, if x and 2x are coloured the same, we are done. Also, this does not affect Schur's theorem later. Do you agree?
at 9:00 (and similar following steps) What if let's say
x1 = 2 x0
then x1-x0 = x0
and the difference has blue color. But there is no triplet because the three numbers are in fact only two numbers.
With the same setup, the last remaining color could have two numbers as well... unless somehow this constellation cannot occur?
Hi @deinauge7894, That's an excellent question, and I missed this subtlety. But going over the proof again, I see that the numbers x,y, and z in Schur's lemma need not be distinct. In other words, if x and 2x are coloured the same, we are done. Also, this does not affect Schur's theorem later. Do you agree?
@MudithaMath Many thanks! I missed the conclusion for this case. Of course I agree.
If it were not a failed attempt, the elegance would've surely earned this proof a place in The Book!
Yeah, definitely!
Diophantine equation:
aⁿ + bⁿ + cⁿ = dⁿ for n>4
is more interesting today. The Fermat is a thing of the past