The equation $x^2 = 2$ has $4$ different solutions in a commutative ring with unity ($1 \neq 0$). Show that in this ring $\exists$ zero divisor. 
The equation $x^2 = 2$ has $4$ different solutions in a commutative
  ring with unity ($1 \neq 0$). Show that in this ring $\exists$ zero
  divisor.

Any help would be appreciate.
I'm only have the following:


*

*A zero divisor is an element $a \in R\ \text{if}\ \exists b\in R, b \neq 0: ab = 0$.

*Well, the equation of the problem can be written as $x^2 -2 = 0
   \implies (x-\sqrt{2})(x+\sqrt{2})=0$
I'm beginning with ring so I'm clueless here. Aside of the question above, how does it possible for $x^2 = 2$ to have $4$ solutions instead of (the known/obvious) two?
 A: Here's the intuition.  You might expect to be able to factor $x^2-2$ as $(x-\sqrt{2})(x+\sqrt{2})$.  If your ring has no zero divisors, the only way a product can be $0$ is if one of the factors is $0$, so the only way $x^2-2$ can be $0$ is if $x-\sqrt{2}=0$ or $x+\sqrt{2}=0$.  That is, the only possible solutions to $x^2=2$ are $x=\sqrt{2}$ and $x=-\sqrt{2}$.  So there can be at most two solutions.
Now to be clear, this argument is nonsense as written.  The symbol $\sqrt{2}$ has no meaning in a general commutative ring.  To make the argument rigorous, you might let $a$ be one of the solutions to $x^2=2$, so you can think of $a$ as a square root of $2$.  Then if you can prove that $x^2-2=(x-a)(x+a)$ for all $x\in R$, the argument goes through.  I'll leave verifying the details to you.
As an example for how $x^2=2$ can have more than two solutions, consider the ring $\mathbb{R}\times\mathbb{R}$.  The element $2$ of this ring is the ordered pair $(2,2)$ (since $2$ really means the sum of two copies of the multiplicative identity, and the multiplicative identity is $(1,1)$).  This has four different square roots: $(\sqrt{2},\sqrt{2}),(\sqrt{2},-\sqrt{2}),(-\sqrt{2},\sqrt{2}),$ and $(-\sqrt{2},-\sqrt{2})$.
A: For example in the ring $R = \mathbb{Z}/119\mathbb{Z}$ you have $11^2 = 121 \equiv 2 \pmod {119}$ and hence $(-11)^2 \equiv 2 \pmod {119}$. But you also have $45^2 = 2025 = 17 \cdot 121 + 2 \equiv 2 \pmod {119}$. Thus you have 4 solutions to $x^2 = 2$: $11, 45, 74 (= -45)$ and $108 (= -11)$.
One way to do this is to note that if $R$ has no zero divisors, then we can form the fraction field $\mathbb{Q}(R)$ and in fields, there can only be two solutions to $x^2 = 2$ (polynomials of degree $n$ have at most $n$ roots in fields).
A: Let $a$ be solution of equation $x^2=2$ then $-a$ is also solution. Since there are 4 solutions then there is $b\ne\pm a$ s.t. $b^2=2$. Then $0=a^2-b^2=(a-b)(a+b)$ and $a\pm b\ne 0.$   
