If $\sqrt{n}$ is not an integer, are there any rational numbers $x$ such that $x^2 = n$? [duplicate]

Question

I have a feeling this has something to do with prime numbers, but I'm totally lost. I'm not sure how to write the proof that proves (or disproves) the following: "Let $n$ be an integer. If $\sqrt{n}$ is not an integer, then there is no rational number $x$ such that $x^2 = n$."

Answer 1

Proof by Contradiction: Let's assume $x$ is rational but not an integer when defined as $x=p/q=\sqrt n$ where $p,q,n$ are integers and $q\ne0$.

$p$ and $q$ can be expressed as products of non-negative integral powers $(p_1,p_2,...,p_n)$ and $(q_1,q_2,...,q_n)$ of prime factors $(a_0, a_1, ... , a_n)$ where $a_n$ is the largest of all the prime factors of $p$ and $q$

$$p=a_0^{p_0}a_1^{p_1}a_2^{p_2}...a_n^{p_n}$$ $$q=a_0^{q_0}a_1^{q_1}a_2^{q_2}...a_n^{q_n}$$

Since, $x$ is not an integer, $q$ does not divide $p$. Hence, $$\exists a_i. q_i>p_i ......... (1)$$

Now, $n=x^2=p^2/q^2$ is an integer. But, $$p^2=a_0^{2p_0}a_1^{2p_1}a_2^{2p_2}...a_n^{2p_n}$$ $$q^2=a_0^{2q_0}a_1^{2q_1}a_2^{2q_2}...a_n^{2q_n}$$

And, from $(1)$, we deduce that $$\exists a_i, 2q_i>2p_i$$

This implies that $q^2$ does not divide $p^2$ which, in turn, implies that $n$ is not an integer. This is a contradiction.

Therefore, the assumption is false and $x$ has to be an integer whenever it is rational. In other words, if $x=\sqrt n$ is not an integer then, it cannot be rational.

Note: With regards to the question "Are there any rational numbers $x$ such that $x^2=n$?", it is easy to prove that for every natural number $n$, there exists a unique real number $x$ such that $x^2=n$. And this unique number is proven to be irrational by the above proof.

Answer 2

Let $n$ be a number such that $\sqrt{n}$ is not an integer. If the equation $x^2-n=0$ has rational roots, the roots are integers. Then $\sqrt{n}$ is an irrational number.