Question

suppose that \(a,b,n \in \mathbb{N}\)\[\gcd(a,b)=1\]and\[ab=n^2\]a,b must be some perfect squares
"and a quite simpler method offerd by @eliassaab down there"

Answer 1

using prime factorization of \(a\) and \(b\)\[a=\prod p_i^{\alpha_i}\]\[b=\prod p_j^{\alpha_j}\]\(p_i\neq p_j\) since \(\gcd(a,b)=1\)

Answer 2

Notice that all of the prime factors of \(n\) can be found in the prime factors of \(a\) and \(b\)

Answer 3

so we can write\[n=\prod\prod p_i^{\beta_i}p_j^{\beta_j}\]\[n^2=\prod\prod p_i^{2\beta_i}p_j^{2\beta_j}\]\[ab=n^2\]this implies that\[ab=\prod p_i^{\alpha_i}\prod p_j^{\alpha_j}=\prod\prod p_i^{\alpha_i}p_j^{\alpha_j}=\prod\prod p_i^{2\beta_i}p_j^{2\beta_j}=n^2\]so\[\prod\prod p_i^{\alpha_i}p_j^{\alpha_j}=\prod\prod p_i^{2\beta_i}p_j^{2\beta_j}\]comparing exponents of both sides gives \[\alpha_i=2\beta_i\]\[\alpha_j=2\beta_j\]

Answer 4

so\[a=\prod p_i^{2\beta_i}=(\prod p_i^{\beta_i})^2=m^2\]\[b=\prod p_j^{2\beta_j}=(\prod p_j^{\beta_j})^2=k^2\]

Answer 5

after all that i must say this is not well written :)

Answer 6

it's a cool way of sarting your green days though

Answer 7

starting*

Answer 8

yup

Answer 9

Congratuations for getting my favorite color @mukushla

Great tutorial I must say..

Answer 10

thank u mere yaar

Answer 11

Welcome benim yoldashim ..

Answer 12

How about this argument?
Suppose that \(a \) is not a perfect square, then there is
a prime p such that \( p^{2n +1}\) is part of the unique prime factorization of \(a \) and gcd(p,b)=1. Hence \( p^{2n +1}\) is part of the unique prime factorization of
\( n^2 \) which is not possible.

Answer 13

thank u sir