<s>[INST] What is the problem here (all integers are irrational proof...I think so)? Let us assume $a$ is an integer which is rational which implies $a=p/q$ (where $p$ and $q$ are integers and $q$ not equal to $0$). If $p$ and $q$ are not coprime, let us simplify the fraction so this it is (I don't know how to talk like mathematicians).
Which implies, $$a=b/c$$ (where $b$ and $c$ are coprime integers). Squaring on both sides,
\begin{align}
a^2&=b^2/c^2\\
a^2c^2&=b^2
\end{align}
So $a^2$ is a factor of $b^2$, and also of $b$, due to the uniqueness of the fundamental theorem of arithmetic.
So,
\begin{align}
b &=a^{2}d   \tag{where $d$ is an integer}\\
b^2 &= a^{4}d^{2}
\end{align}
But $b^2=a^2c^2$ So,
\begin{align}
a^2c^2 &= a^4d^2\\
c^2 &= a^2d^2
\end{align}
So, $a^2$ is a factor of $c^2$ and $c$ due to the fundamental theorem of arithmetic. So $b$ and $c$ have $a^2$ as a common factor. But this contradicts the fact that $b$ and $c$ are coprime. This is because we have taken $a$ as a rational integer, so $a$ cannot be a rational integer.
What's wrong here (genuinely asking)?
 [/INST] I think you are confusing that if $p$ is prime and $p$ divides $b^k$ then $p|b$.  That is true if $p$ is prime.
Actually it's also true for a composite $a|b^k$ then $a|b$ if $a$ has no square factors.  But if $a$ as any prime factors to a power greater than $1$ it need not be true.  
And in fact its obviously not true as $a^2$ divides $a^2$ but $a^2$ doesn't divide $a$ (unless $a = 1$). 
Read on....
It most certainly is not true if $a|b^k$ that $a|b$   It means that the prime factors of $a$ are prime factors of $b$.  And it means that the powers of those prime factors of $a$ are at most equal to $k$ times the powers of the same prime factors of $b$ but because $k$ is larger than .....
Oh let me put it this way.
Suppose $a = \prod p_i^{m_i}$ be the prime factorization of $a$.  Suppose $a|b^k$.  Then that means that $p_i$ are prime factors of $b$ and that $b = d\prod p_i^{j_i}$.  And it means that $b^k = d^k \prod p_i^{k*j_i}$.
And as $a|b^k$ that means each $m_i \le k*j_i$.  But that does not mean $m_i \le j_i$ which would mean $a|b$.
You statement $a|b^k$ means $a|b$ if $a$ has square free and all the prime factor powers were $1$ but not other wise.
Simple example if $a = 12 = 2^2*3$ and $b= 90 = 2*3^2*5$.  Now $a|b^2 = 8100 = 2^2*3^4*5^2$.
This means the prime factors of $a$ ($2,3$) are also prime factors of $b$.  And it means that the powers of the prime factors of $a$  ($2\mapsto 2; 3\mapsto 1$) are less or equal to $2$ times the powers in $b$  ($2\mapsto 1$ and $2 \le 2*1$ and $3\mapsto 2$ and $1 \le 2*2$) but it doesnt mean the are less than or equal to the powers of $b$.  (In $a; 2\mapsto 2$ but in $b; 2\mapsto 1$ and $2 \not \le 1$).
So $12 \not \mid 90$.
It's certainly can't be the case that $a|b \implies a^2| b^2 \implies a^2|b$!  That would mean every time you have $a|b$ you can just keep squaring and reducing to get $a^{m}|b$ for any power of $m$.
That would mean if $3|6$ then $3^2|6$ and $3^4|6$ and $3^{2048}|6$ and so on.
Or in this case as $a = b$ (and $c=1$.... because $a$ is an integer) you would have $a|a$ so $a^2|a$?  And $a^4|a$.  That's .... simply not true.
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