If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$? The common multiplication rules are 
$$(-1) \cdot (-1) = +1 \\ 
(+1) \cdot (+1) = +1$$
But these rules seem asymmetric. Because of these rules it is not possible e.g. to solve the equation
$$x^2 = -1 $$
without introducing complex numbers. There would be two "more symmetric" alternatives:
$$(-1) \cdot (-1) = -1 \\ 
(+1) \cdot (+1) = +1$$
or
$$(-1) \cdot (-1) = +1 \\ 
(+1) \cdot (+1) = -1$$
Both alternatives would allow to solve the equations
$$x^2 = +1 \\ x^2 = -1 $$
without the need to introduce complex numbers. However the two more symmetric alternatives are not used. What is the reason for that? Would math break down, if one would adopt one of these different conventions?
 A: A lot of math would break down if you use that, yes. For example, with our multiplication, we have the property that for any three numbers $a,b,c$, we have $$a(b+c) = ab + bc$$
This would not hold anymore in your case, since
$$0 = -1\cdot 0 = -1\cdot (1 + (-1)) \neq 1\cdot (-1) + (-1)\cdot (-1) = -1 -1 =-2$$
This is assuming that the new multiplication is still commutative and that $1$ is still the "unit" of multiplication, i.e. that $1\cdot x = x$ for all $x$.

In fact, all you need to prove that $(-1)\cdot (-1) = 1$ is the rule $a(b+c) = ab+bc$, and the property that $1\cdot x = x$, since you can see that if this rule is true, then
$$0 = (-1)\cdot 0 = (-1)\cdot (1+(-1)) = (-1)\cdot 1 + (-1)\cdot(-1) $$
which means that $0=(-1)\cdot (-1) + (-1)\cdot 1$ or in other words that $(-1)\cdot (-1) = 1$.
A: In some branches we do have that $1\cdot 1=-1$. However, the role of $1$ is by definition that of a neutral element of multiplication (like $0$ is neutral for addition), i.e., $1\cdot x=x\cdot 1=x$ holds (or is postulated to hold) for all $x$. So specifically $1\cdot 1=1$. This does not immediately contradict your idea of $1\cdot 1=-1$, it just implies $1=-1$ or in other words $2=0$. This equation does hold in the field $\Bbb F_2$ of two elements or more generally in fields of what is called characteristic $2$, which are widely applied for example in computer science and coding theory and other areas.
Apart from this way of sneaking away from the original question, it may be that your main problem comes from the introductory words "The common multiplication rules are..." In fact, what you cite then are not the common basic rules (or axioms) of multiplication, which are instead
$$ a\cdot(b\cdot c)=(a\cdot b)\cdot c,\qquad (a+b)\cdot c=(a\cdot c)+(b\cdot c),\quad a\cdot(b+c)=(a\cdot b)+(a\cdot c).$$
We can infer the rest from these. Specifically, the properties of addition force us to accept that $0\cdot a=a\cdot 0=0$ for all $a$: From $0+0=0$ and distributivity we see that $0\cdot a=(0+0)\cdot a=0\cdot a+0\cdot a$, hence $0\cdot a=0$ and similarly for the other order. As a consequence of this, we have $a\cdot b+a\cdot c=0$ whenever $b+c=0$, i.e. $a\cdot(-b)=-(a\cdot b)$ (and similarly $(-a)\cdot b=-(a\cdot b)$). So at any rate $(-a)\cdot(-a)+a\cdot(-a)=0$ and $a\cdot a+a\cdot (-a)=0$ and consequently $$(-a)\cdot(-a)=a\cdot a.$$
A: The rules $(-1)\cdot(-1)=1$ and $1 \cdot 1 = 1$ aren't conventions.  They are consequences of how we define our number system: specifically, of what multiplication is, and what the symbol $-1$ means.
You can, if you like, define brand new operations, with different symbols, obeying different rules, and study the properties of such systems.  That is in fact what group theory is all about.
A: Here is a very "naive" way of answering this question.

When trying to define the following:


*

*Positive $\times$ Positive $=$?

*Positive $\times$ Negative $=$?

*Negative $\times$ Positive $=$?

*Negative $\times$ Negative $=$?


We first need to ensure that the answers to #2 and #3 are identical.
This is because multiplication is commutative ($\forall a,b:a\times b=b\times a$).

Let's arbitrarily define:


*

*Positive $\times$ Negative $=$ Negative

*Negative $\times$ Positive $=$ Negative


As you can see:


*

*In the output, the sign of the positive input has been inverted

*In the output, the sign of the negative input has been preserved


Following this "behavior", the answers to #1# and #4 are clear:


*

*Positive $\times$ Positive $=$ Positive

*Negative $\times$ Negative $=$ Positive



Of course, we could always arbitrarily define:


*

*Positive $\times$ Negative $=$ Positive

*Negative $\times$ Positive $=$ Positive


And subsequently:


*

*Positive $\times$ Positive $=$ Negative

*Negative $\times$ Negative $=$ Negative


But that would only make a semantic swap between "positive" and "negative".
A: Math would certainly break down, here's why. Assume that every number is equal to itself, addition is associative, multiplication distributes over addition , $1$ and $-1$ are additive inverses, $1$ is the multiplicative identity, and $0$ is the additive identity. Then it follows that:
$$\begin{array}{lll}
\bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1) &=& \bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1)\\
\bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + \bigg(1\cdot (-1) + (-1)\cdot(-1)\bigg)\\
1\bigg(1+(-1)\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + (-1)\bigg(1 + (-1)\bigg)\\
1\bigg(0\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + (-1)\bigg(0\bigg)\\
0 + (-1)\cdot(-1) &=& 1\cdot 1 + 0\\
(-1)\cdot(-1) &=& 1\cdot 1 \\
\end{array}$$
The only loose end not assumed above is  $(-1)\cdot 0 = 0$
$$(-1)\cdot 0 = (-1)(0+0)$$
$$(-1)\cdot 0 = (-1)\cdot 0+ (-1)\cdot 0$$
$$\dots$$
