1) "Since a<b are integers then sqrt(q) = m/n where m and n are coprime integers" ... that's what you said, so I'm wondering how you got "since a<b" ... and I got what you meant by coprime :)
2) b = ka + r ... if a>b, then k = 0 (my mistake

and r = b ... and from "r = a(sqrt(q) - k)" which Furry got, we get that r = a*sqrt(q) > a, so we still get the contradiction :).
3) Could you show how if the a's cancel out, it would contradict the premise that a^2|b^2?
This is what I get:
As you said, n|a, and if a|n => a = n => b^2 = m^2 => b = m => a = 1. I don't see any problems with a = 1, although it is a trivial case ...
4) By definition of division .. X|Y iff there exists and integer Z such that XZ = Y
So, if a^2|b^2, then there exists an INTEGER q such that qa^2 = b^2. If sqrt(q)=m/n (reduced fraction), then q = m^2/n^2 ... which is still a reduced fraction, so it's a rational number, not an integer, unless n = 1.
Yes, if n|a, then b is still an integer, but q itself is not an integer anymore (unless n = 1), so it still doesn't work unless n = 1.
5) I agree with your statement "
The answer is that if n must divide a but if n were greater than 1 that would contradict the premise that a^2|b^2", that also works out, but for me, the reason n = 1 is because sqrt(q) is also an integer (so if we consider the set of rationals, then q is in the subset of the rationals in which the denominator is 1).
Please remember, this is in reference to this statement that you made:
"
since r = a(sqrt(q)-k) = b-ak we have a*sqrt(q) = b. Since a<b are integers then sqrt(q) = m/n where m and n are coprime integers.
am=nb. But "a" does not divide n because if it did then the a's cancel out in the formula b^2 = a^2m^2/n^2. The only answer is that "a" must divide b. Or is that based on the fundamental theorem of arithmetic again."
Here's an example ...
36^2 = 1296 = (4^2 / 3^2) * 27^2 ... there, that example should work ... b = 36, a = 27, m = 4, and n = 3). I believe all your conditions are satisfied, n|a, m and n are relatively prime, and a does not divide n. Does a|b (27|36)?
My main problem is that in your proof, you did NOT indicate that n=1 (you just said that a does not divide n, which can be true), and the reason I find this to be a problem is that it contradicts the fact that q is an integer. That's how I get n=1, because if n is not 1, then q is not an integer. Another minor issue is that you did not include the special case where a|n and a|b, this is when a = n = 1, but again, you did not mention anything that says there would be a problem if n|a, with n>1.
6) Yes, my previous post did invoke the FTA, but just as another way you can look at it. Aside from that, I don't think the FTA needs to be invoked outright.
Example:
a and b are integers such that a^2|b^2 => ma^2 = b^2
Since the square root of b^2 = b, then we can take the sqrt(ma^2) = a*sqrt(m) = b.
Since a and b are integers, sqrt(m) must also be an integer
=> a|b
I don't see the FTA there ... and you could probably use the same idea for a^x|b^x => a|b
Furry, if you want, you can ask your teacher if he/she would find the above proof to be acceptable :) ... and if it's not, please let me know what he/she thinks is wrong with it :)