mr-mitch wrote:The intuition (well maybe not intuition) is something like if you have am² and bn², then they're basically a product of primes, and the list of primes is the same. If m and n are non-trivial products of primes and m and n share no common primes, the remaining primes must come from a and b, and so a and b are square.
We were told this was not the case, and therefore the prime factorisation of m and n is trivial.
yyy wrote:It is also true that |a|=|b| for the same reason, though they are not necessarily equal to 1.
One of the known proofs that the square root of 2 is irrational reaches contradiction through2m^2=n^2
gorcee wrote:Ah, I see it now. m^2 has some prime factorization p_1^2 p_2^2... which cannot be in n^2, so it must be in b. Gotcha.
MartianInvader wrote:gorcee wrote:Ah, I see it now. m^2 has some prime factorization p_1^2 p_2^2... which cannot be in n^2, so it must be in b. Gotcha.
Actually, you should replace those "2"'s with "2i" where i is any integer (a fourth power is still a square!)
That whole extra assumption about the gcd seems strange to me. Can't you just say that |a|=|b| and |m|=|n| instead?
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