Prove that $x^2 = - 1$ has no solution in $\mathbb{Z}$ I have this proposition to prove: The equation $x^2 = -1$ has no solution in $\mathbb Z$.
I was told that this is an opportunity for a proof by contradiction. I have already proven that for $m \in\mathbb Z$, if $m \ne 0$ then $m^2 \in\mathbb N$. I have also proven that 1 is a natural number (hence, -1 isn't one).
Here is my strategy: assume that there is a solution. Hence $x^2$ has 3 options:
\begin{align*}
x^2 \in\mathbb N\\
-(x^2) \in\mathbb N\\
x^2 = 0
\end{align*}
Option 1 is not possible because -1 is not a natural number. Option 2 doesn't make sense because       
-$\mathbb N \in \mathbb N$ and option 3 doesn't make sense either because $0 \ne -1$. What do you think? Thank you!

Here is a simpler strategy based on the comments below.
Proof: Assume that $x^2 = -1$ has a solution in $\mathbb Z$, then $x^2$ will either be $0$ or $\in\mathbb N$. However, $-1 \notin\mathbb N$ and $-1 \ne 0$. Hence, there is a contradiction. Here is how I have proven that $-1 \notin\mathbb N$: Let $m \in\mathbb N$. Hence:
\begin{align*}
-m \notin\mathbb N\\
(-1)m \notin\mathbb N
\end{align*}
If $-1 \in\mathbb N$, the product of $-1$ and $m$ should $\in\mathbb N$, which is not the case here. Hence, $-1 \notin\mathbb N$. 
 A: I don't think your solution makes sense. There's definitely something circular going on when you handle the second case - and I'm not sure how the structure of your proof was intended. It would be better to split into cases based on $x$, rather than $x^2$. In particular, a proof of this could run as follows:


*

*Suppose $x\geq 0$. Then $x^2\geq 0$. However $-1$ is not greater than or equal to $0$, and hence $x^2$ cannot be $-1$.

*Suppose $x\leq 0$. Then $x^2\geq 0$. Similarly as before, $x^2$ cannot be $-1$.


However, given that you've already shown that $x^2\in \mathbb N$ if $x\neq 0$, you already know that if $x\neq 0$ then $x^2\neq -1$ because $-1\not\in \mathbb N$. Since $0^2$ is also not $-1$, the theorem is proven.
A: Any integral solution whose magnitude is greater than one is impossible, because the squared term would be too large. So you only have to check $\pm1$, and both fail.
A: An alternative approach is to observe that $z^2+1=0$ has at most two complex solutions since the number of zeros of a polynomial is less than or equal to its degree. $i,-i$ are two such solutions so there are no others.  Neither of these is a natural number so there are no natural number solutions. 
I call this approach "extension-uniqueness."
A: I will attempt to prove this by contradiction here.
Suppose instead there does exist a solution in $\mathbb{Z}$ to the equation $x^2=-1$. Algebra would tell us that the only solutions to that equation are $x=\pm i$. However, $i$ and $-i$ are both complex numbers. So $i$ and $-i$ are not real numbers, let alone not integers. (Note that $\mathbb{Z} \subset \mathbb{R}$.) So $i,-i \not\in \mathbb{Z}$. This is the contradiction.
A: The mos straight-forward technique is to show that $x^2+1$ is irreducible over $\mathbb{R} \Rightarrow$ irreducible over $\mathbb{Z}$. For a degree two polynomial $f(x)$ to be reducible it must have a liner factor, i.e a root $u$ satisfying $f(u)=0$. Here by the quadratic formula we know $x^2+1 = 0 \Rightarrow$
$$x = \pm \dfrac{\sqrt{0^2-4}}{2} = \pm \dfrac{\sqrt{-4}}{2} = \pm \sqrt{-1} = \pm i$$
Hence if $x^2+1$ were to factor it would do as the product $(x+i)(x-i)$, where neither $\pm i \in \mathbb{Z}$ (since if so the integers would be an ordered field).
A: Suppose there is a solution, then reduce $\mod 4$. There are no solutions to $x^{2} \equiv 3 \mod 4$ since the only squares $\mod 4$ are, $0$ and $1$. Thus you have a contradiction. But, this requires more work than Bill's solution. His solution only requires you to check 3 numbers. 
A: Since 0^2+1=0+1=1(not equal to 0) ; then x€real number is not a solution of x^2+1=0
therefore x^2+1=0 has a non-zero solution.
Now we have to extablish the theorem by method of cotradiction. 
Let the equation has real solution. 
This is sufficient to prove that of any non-zero real number which is a solution of the equation. 
On its contrary. spose there exist x belongs to real numbers - {0} as a solution of the equation.
But x not equal to 0
Implies x^2>0
Implies x^2+1>0
For all x belongs to zero and x not equal to zero 
Imples x^2+1 not equal to zero
Which is a contradiction. 
So our assumption is wrong. 
As a result x^2+1=0 has no real solution. 
