> Why not? It seems at this point you are assuming something that is generally deduced as a consequence of the FTA.
According to the OP, this result does not rely on the FTA -- he claims to derive it from the Euclidean Algorithm. makomk does the derivation in a sibling comment.
> In particular, you have assumed that the product of the p_m is k (with appropriate exponents), and because c is in p_n we know that c|k, and hence we know that k=0 (mod c). So your claim here is false.
I don't see any problem there? I say that k = 0 (mod c) because c is in p_n, and k ≠ 0 (mod c) because c is not in p_m. That's a contradiction, which is what I wanted to show. The remaining possibility is that p_n and p_m contain the same prime factors in different quantities, and the rest of the proof reduces that case to this same contradiction.
EDIT -- ColinWright has quoted text which I edited out of this comment; his quote is accurate.
>> Why not? It seems at this point
>> you are assuming something that is
>> generally deduced as a consequence
>> of the FTA.
> According to the OP, this result
> does not rely on the FTA -- he
> claims to derive it from the
> Euclidean Algorithm.
Yes.
> I can't do that, so I'm taking
> his word for it, but that doesn't
> make the proof circular.
But you should say that you are relying on this. As it is you are simply making an unsupported assertion, and so your proof is incomplete.
See other comments in this sub-thread for more explanations.
According to the OP, this result does not rely on the FTA -- he claims to derive it from the Euclidean Algorithm. makomk does the derivation in a sibling comment.
> In particular, you have assumed that the product of the p_m is k (with appropriate exponents), and because c is in p_n we know that c|k, and hence we know that k=0 (mod c). So your claim here is false.
I don't see any problem there? I say that k = 0 (mod c) because c is in p_n, and k ≠ 0 (mod c) because c is not in p_m. That's a contradiction, which is what I wanted to show. The remaining possibility is that p_n and p_m contain the same prime factors in different quantities, and the rest of the proof reduces that case to this same contradiction.
EDIT -- ColinWright has quoted text which I edited out of this comment; his quote is accurate.