Why is it impossible to find natural numbers $a$ and $b$ such that $12b^2=a^2$? This was a question in the exercises for an EdX course by Prof Starbird on Effective Thinking through Mathematics which was long over, but I am working through the course at my own pace.
I feel that the 'correct' answer given by the course is wrong - I do not want to paste it here because it may violate the Course Agreement (even though the course ended 3 months ago).
"Why is it impossible to find natural numbers $a$ and $b$ such that $12b^2=a^2$?"
Can you please look at my reasoning and tell me what is missing?
My attempt:
Assume it is possible to find natural numbers $a$ and $b$ such that $12b^2=a^2$
Then
$$3 = \frac{a^2}{2^2\cdot b^2}$$
Implies that $a^2$ is even. The prime factorization of $a^2$ will contain $2\cdot 2$.
Assume $a = 2\cdot c$ where $c$ is odd.
So
$$3 = \frac{c^2}{b^2}$$
Now $b^2$ is also odd. But it does not mean that the prime factors of $b$ and $c$ contain $3$.
The 'correct' answer seems to assume that the prime factors of $b$ and $c$ contain $3$.
I feel this is incorrect. 
 A: In your proof, you can set $a=2c$, but note that $c$ may be even. Note that all we know is that $a$ has at least one $2$ as its prime factor. (You may miss the case when $b$ is even.)
We can prove the statement in the following way :
For $2^2\cdot 3\cdot b^2=a^2,$ the number of the prime factor $3$ in LHS is odd while the number of the prime factor $3$ in RHS is even. This is a contradiction. Q.E.D.
A: That's a variation on the classical proof that $\sqrt{2}$ is not rational.
If $12b^2=a^2$, you can assume $a$ and $b$ have no common factor, otherwise you factor them out. Then $3$ divides $a^2$, but $3$ is prime, so it divides $a$. Then $a=3c$, so $12b^2=9c^2$ thus $4b^2=3c^2$, and $3$ divides $b$. Contradiction, so your equation has no solution.
A: HINT 
Any perfect square number must contain even powers in its prime factorization :
$$n^2 =  p_1^{e_1}p_2^{e_2}\ldots p_r^{e_r}$$
$$e_i =  2k$$
A: $3$ is divisible by $3$ and has prime factorization $1 \cdot 3^1$.
Therefore, if $3 = b^2 / c^2$, then the $b^2 / c^2$ is divisible by $3$, and has prime factorization $1 \cdot 3^1$.
