Recent content by rerouter

Yes, I think that makes sense. I thought that is what I had shown, "sort of", lol. For some reason I think I wrote the same thing twice with different variables. So let's say n=pj, p and j are primes. (This is the prime factorization of n). Since n|a2 and n|b2, we know that. a2=(pj)s...

Maybe. So let's say n=pj, p and j are primes. a2=pjs, p, j, and s primes. b2=pjr, p, j, and r primes. n|a2 means that a2=nw a2=pjv n|b2 means that b2=nx b2=pjy Since a=(pj)k(pj)1/2x1/2 b=(pj)k(pj)1/2y1/2 a and b are multiples of pj, so n|ab ?

If I consider n=pks and n|a2 then a2=nt a2=pkst It looks like a=sqrt(pkst) a=pk/2sqrt(st) so that is k/2 p's ?

I think then: a and b would then each have at least k factors of p in them.

Forgive me for being dense, lol.

While I see that that's true...I'm not sure how to show that. Given: m2|a2 a2=m2k (...k some integer) so.. a=m*sqrt(k) Showing that a is a multiple of m, thus m|a. But how can I justify this? What if sqrt(k) is not an interger. I know from a few trials, that k either equals...

Hmmm... m<=a ?

Thanks for the welcome tiny-tim, I appreciate it! I just looked at it again using n2 instead of n. I still have the problem that I can't seem to isolate ab. An example: a2=n2m b2=n2k Lets say I multipy those by b and a respectively, to try and get some ab's. a2b=n2mb b2a=n2ka...

Homework Statement Question: If n|a^2 and n|b^2 prove or disprove that n|ab. I think this is true, but am having trouble proving it. Homework Equations a^2=nm, b^2=nk a^2-b^2=n(m-k) a^2+b^2=n(m+k) The Attempt at a Solution Essentially I've tried all sorts of algebraic...