Squarefree number property

[gorcee]

This is not homework, merely a curiosity I came across.

I encountered the statement "If am²=bn², where a,b are squarefree integers and m,n are integers such that gcd(m,n) = 1, then |m|=|n|=1."

However, I am having difficulty seeing the reasoning behind this statement.

[mr-mitch]

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.

[gorcee]

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.

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.

[yyy]

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 through

2m^2=n^2

[Snark]

Let a's prime factorization be a₁*a₂*...*a_p where a_i =/= a_j for i =/= j
Let b's prime factorization be b₁*b₂*...*b_q where b_i =/= bj for i =/= j

Let am² = bn²
Let gcm(m,n) = 1
Let m's prime factorization be m₁*m₂*...*m_s
Let n's prime factorization be n₁*n₂*...*n_t

am² = a₁*a₂*...*a_p*m*m
bn² = b₁*b₂*...*b_q*n*n
If am² = bn²
then a₁*a₂*...*a_p*m*m = b₁*b₂*...*b_q
then a₁*a₂*...*a_p*m₁²*m₂²*...*m_s² = b₁*b₂*...*b_q*n₁²*n₂²*...*n_t²
then if m₁² > 1, then no subset of factors on the right can multiply to equal m₁² since a_i =/= a_j for i =/= j and gcd(m_i, n_j) = 1
thus m₁² = 1, Thus m₁ = 1
By similiar proof all m_i's = 1 and all n_i's = 1
then m = 1 and n = 1
Since m=n=1, a=b


Ta-da!
Also: I hope I never have to hit "sub" or "sup" again.

[gorcee]

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 through

2m^2=n^2

Right, that I know, it was just the generalization to two squarefree numbers that I had a hiccup on.

[MartianInvader]

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?

[gorcee]

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?

The assumption from the gcd = 1 comes from the fact that we're dealing with elements of Q(a), so what we actually have is a situation where a = b q^2, and we let q = m/n, where gcd(m,n)=1 WLOG.