Prove that $\sqrt{n^2 + 2}$ is irrational 
Suppose $n$ is a natural number. Prove that $\sqrt{n^2 + 2}$ is irrational.

From looking at the expression, it seems quite obvious to me that $\sqrt{n^2 + 2}$ will be irrational, since $n^2$ will be a natural number, and after adding $2$ to it, $n^2 + 2$ will no longer be a perfect square.
From that logic, I tried letting $n^2 + 2$ = ${a^2\over b^2}$ to find a contradiction, but I wasn't able to formulate a proof that worked. Any suggestions as to how I should approach this?
 A: It can be easily proven that $\sqrt k$ is irrational if $k\in$ N is not a perfect square. At the same time, the difference between two consecutive perfect squares is $(n+1)^2-n^2=2n+1>2$ for $n>0$. So $n^2+2$ lies in between two consecutive perfect squares, meaning it is not one. Therefore, its radical is irrational.
A: Without loss of generality we can assume $x,y\in \mathbb{N}\ (?)$.
$$n^2+2=\dfrac{x^2}{y^2} \implies x^2=y^2\left(n^2+2\right)$$
For odd or even $n$, $x,y$ can be both odd, or exactly one of them should be even. 
Case 1 ($n$ odd)
Case 1.1 (both odd)
$\left(x^2\equiv1\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv3\pmod 4$
Case 1.2 ($x$ odd, $y$ even)
$\left(x^2\equiv1\pmod4, y^2\equiv0\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right)  \implies 1\equiv0\pmod 4$
Case 1.3 ($x$ even, $y$ odd)
$\left(x^2\equiv0\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right)  \implies 0\equiv3\pmod 4$
Case 2 ($n$ even)
Case 2.1 (both odd)
$\left(x^2\equiv1\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv2\pmod 4$
Case 2.2 ($x$ odd, $y$ even)
$\left(x^2\equiv1\pmod4, y^2\equiv0\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right)  \implies 1\equiv0\pmod 4$
Case 2.3 ($x$ even, $y$ odd)
$\left(x^2\equiv0\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right)  \implies 0\equiv2\pmod 4$
A: When $n$ is even, show that $n^2+2$ will be $2$ times an odd number. Then show that such a number doesn't have a rational square root (you can modify the usual proof that $\sqrt{2}$ is irrational).
When $n$ is odd, show that $n^2+2\equiv 3$ (mod $4$). Then show that any integer squared is congruent to $0$ or $1$ mod $4$, so that $n^2+2$ is not a perfect square. As in Jaoa's comment, finish up by arguing the square root of a non-square is irrational, which again can be done by modifying the $\sqrt{2}$ proof.
A: $$\begin{array}{l}
 \sqrt {2 + n^2 }  = \frac{p}{q};\quad and\quad p \wedge q = 1 \\ 
 \sqrt {2 + n^2 }  = \frac{p}{q} \Leftrightarrow 2 + n^2  = \left( {\frac{p}{q}} \right)^2  \\ 
  \Leftrightarrow 2 = \left( {\frac{p}{q}} \right)^2  - n^2  \\ 
  \Leftrightarrow 2 = \left( {\frac{p}{q} - n} \right)\left( {\frac{p}{q} + n} \right) \\ 
  \Leftrightarrow 2q^2  = \left( {p - nq} \right)\left( {p - nq} \right) \cdots \left( * \right) \\ 
 \left( * \right) \Rightarrow \left\{ \begin{array}{l}
 \exists m_0  \in\mathbb{ N} :p - nq = m_0 q \\ 
 or \\ 
 \exists m_1  \in \mathbb{N} :p + nq = m_1 q \\ 
 \end{array} \right. \\ 
  \Rightarrow \left\{ \begin{array}{l}
 \exists m_0  \in \mathbb{N} :p = \left( {n + m_0 } \right)q \\ 
 or \\ 
 \exists m_1  \in \mathbb{N} :p = \left( { - n + m_1 } \right)q \\ 
 \end{array} \right. \\ 
 \left\{ \begin{array}{l}
 p = \left( {n + m_0 } \right)q \Rightarrow q/p \\ 
 or \\ 
 p = \left( { - n + m_1 } \right)q \Rightarrow q/p \\ 
 \end{array} \right. \\ 
 \end{array}$$
contradiction  with $\quad p \wedge q = 1$
A: If $n\equiv0\pmod{2}$, then $n^2+2\equiv2\pmod{4}$.
If $n\equiv1\pmod{2}$, then $n^2+2\equiv3\pmod{4}$.
Thus, $n^2+2$ cannot be a perfect square since neither $2$ nor $3$ is a quadratic residue mod $4$.
As shown in this answer or using the Rational Root Theorem, if $n^2+2$ is not a perfect square, then $\sqrt{n^2+2}$ is irrational.
