Using the fundamental theorem of arithmetics I’ve got a problem that I’m not quite sure how to solve. I can see the reasoning behind the problem, but I’m not sure how to apply the theorem. 

Suppose that $a$, $b$, and $c$ are integers, and $a^2 = b \cdot c^2. $ Then, $b$ must be a square, e.g. there is some integer $m$ so that $m^2 = b$.

We’re supposed to show this using the fundamental theorem of arithmetic. It might just be because it’s rather late here, but I don’t quite see how the fact that these numbers can be factored into primes helps.
 A: You have to show that, for any prime $p$ in the factorisation of $b$, the exponent of $p$ (what is called the $p$-valuation of $b$) is even.
Indeed the equality $a^2=bc^2$ translates, at the $p$-valuations level,
as
$$v_p(a^2)=2v_p(a)=v_p(bc^2)=v_p(b)+2v_p(c),\enspace\text{whence}\quad v_p(b)=2\bigl(v_p(a)-v_p(c)\bigr).$$
A: Hint. Factor $a$, $b$ and $c$ into primes (in the unique way guaranteed by the Fundamental Theorem of Arithmetic). Then you want to show that  each prime that occurs in $b$ does so an even number of times.  
A: If you consider the prime factorization of $a$, you can see that the prime factorization of $a^2$ will have each prime appearing an even number of times.
Similarly the prime factorization of $c^2$ will have each prime appearing an even number of times.
Then the prime factorization of $b$ must involve the primes in the prime factorization of $a^2$ that do not appear in the prime factorization of $c^2$.
Each prime still appears an even number of times, so $b$ is also a square.
A: Write $a=\prod_ip_i^{A_i},\,c=\prod_ip_i^{C_i}$, where the $p_i$ run over all prime factors of $ac$. Then $b=\prod_ip_i^{2(A_i-C_i)}$, so you can take $m=\prod_ip_i^{A_i-C_i}$. Since $c^2|a^2$, each $A_i\ge C_i$, so this $m$ is an integer.
A: Let $p$ be a prime factor of $b$. Then $p$ is a prime factor of $a$ and therefore, if you express $a^2$ as a product of powers of distinct primes, $p$ will appear there with an even power. But either $p\nmid c$ or $p$ appears in the expression of $c^2$ as a product of powers of distinct primes also with an even power. But then, since $a^2=bc^2$, $p$ appears in the expression of $b$ as a product of powers of distinct primes with an even power too. Since this occurs for every prime factor of $b$, $b$ is a perfect square.
