Recent content by rofler

In the proof, you explicitly use the theorem that there is no string of n consecutive numbers below n^2 so that they are all divisible by at least one of 3,5,...,n, and you must prove this, or it will not be valid. Furthermore, even if CRGreathouse cannot construct a counterexample (although I...

Then you must prove that he cannot create a counterexample. So first and foremost, if you are using the fact that his counterexample does not work because it is larger than n^2, then you must then prove the following statement for your proof to be valid: We cannot have n consecutive...

But what he then proved if it was corrected is really quite simple, isn't it? Take N(S)= lcm(all primes in S), and then if p_1,p_2,...,p_k are the primes in this set that divide n, take m to be 1 mod each of these primes and either 1 or 2 mod the rest, so that m and n-m are relatively prime to...

Ah, I meant n!+3 n!+5, ..., n!+n, but in retrospect, it was meant n consecutive numbers, not n consecutive odd numbers, so chinese remainder would not quite work, sorry (I was thinking of setting up the congruences a=0 mod 3, a+2=0 mod 5, a+4=0 mod 7,... but that was before I realized it was n...

You're missing the point... I know there is a flaw, and it is quite obvious where it is based on what you said... I just wanted to understand what he was TRYING... he's clearly not a mathematician, however I am certainly qualified enough in mathematics to understand every word that you said, so...

Well, could you explain what A1, A2, B1, B2, etc. are? I can't comment on the quality of the proof if I don't understand/can't read/ don't want to turn my head 90 degrees to read the second page. Cheers, Rofler By the way, saying that there is no set of n consecutive numbers UNDER n^2 each...

Hey, CRGreathouse, can you translate his proof into English rather than picture? I have no clue what A1, A2,... B1, B2,... C1, C2,... are, and I believe it would easier to discuss the argument, and possibly any flaws, if it was written like this. Cheers, Rofler