## Squarefree number property

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### Squarefree number property

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.
gorcee

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### Re: Squarefree number property

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.
mr-mitch

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### Re: Squarefree number property

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.

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.
gorcee

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### Re: Squarefree number property

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
yyy

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### Re: Squarefree number property

Let a's prime factorization be a1*a2*...*ap where ai =/= aj for i =/= j
Let b's prime factorization be b1*b2*...*bq where bi =/= bj for i =/= j

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

am2 = a1*a2*...*ap*m*m
bn2 = b1*b2*...*bq*n*n
If am2 = bn2
then a1*a2*...*ap*m*m = b1*b2*...*bq
then a1*a2*...*ap*m12*m22*...*ms2 = b1*b2*...*bq*n12*n22*...*nt2
then if m12 > 1, then no subset of factors on the right can multiply to equal m12 since ai =/= aj for i =/= j and gcd(mi, nj) = 1
thus m12 = 1, Thus m1 = 1
By similiar proof all mi's = 1 and all ni's = 1
then m = 1 and n = 1
Since m=n=1, a=b

Ta-da!
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### Re: Squarefree number property

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 through
2m^2=n^2

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

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### Re: Squarefree number property

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?
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

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### Re: Squarefree number property

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?

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.
gorcee

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