Proving properties of rings None of these are complete and I am looking for some guidance.

  
*
  
*In any ring if $ab=-ba$ then $(a+b)^2=(a-b)^2=a^2+b^2$.
  

So assuming $ab=-ba$ to be true, then squaring both sides gives $(ab)^2=(-ba)^2$, i.e. $abab=-ba-ba$. Not all rings are commutative under multiplication so I don't know if I can assume that $abab =aabb=a^2b^2$.


  
*In any integral domain, if $a^2=b^2$ then $a=\pm b$
  

I'm not positive on how the integral domain plays a part in this one? Something to do with not having zero divisors I'm sure, possibly that no two elements multiplied together will give zero unless one of the elements is zero? So we can just take the square root


  
*In any integral domain, only 1 and -1 are their own multiplicative inverses.
  

We must have a commutative ring with unity, thus we have a multiplicative inverse. Can we assume that 1 and -1 are the inverses?


  
*In any integral domain, if $a^n=0$ for some integer $n$ then $a=0$.
  

Just thinking, the only time anything to a power is zero is when $0^n$ where $n\neq 0$, thus $a$ must be 0.
 A: 1 -
$$ (a+b)^2 = a^2 + ab + ba + b^2, (a-b)^2 = a^2 - ab - ba + b^2 $$
2 -
$$ a^2 - b^2 = (a + b)(a - b) = 0 $$
3 - Particular case of 2 where $b = 1$
4 - Show that $a^2 = 0 \iff a = 0$ and proceed with induction.
Hope this helps.
A: $$(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+b^2$$
$$(a-b)^2=(a-b)(a-b)=a^2-ab-ba+b^2=a^2+b^2.$$
$$(ab)^2=abab=-baab=-ba^2b=a^2bb=a^2b^2$$
$$(a-b)(a+b)=a^2+ab-ba-b^2=a^2+2ab-b^2.$$
$$a^2+a^2=(a+a)(a+a)=a^2+a^2+a^2+a^2.$$
Thus, $2a^2=0$ and from here we can not get $a^2=0$ for all ring. 
A: 1:
$(a + b)^2 = a^2 + ab + ba + b^2 = a^2 + ab - ab + b^2 = a^2 + b^2; \tag 1$
$(a - b)^2 = a^2 -ab - ba + b^2 = a^2 - ab + ab + b^2 = a^2 + b^2; \tag 2$
2:
$(a - b)(a + b) = a^2 - ab + ab - b^2 = a^2 - b^2 = 0, \tag 3$
which works since integral domains are commutative.  Now if 
$a \ne b, \tag 4$
then 
$a - b \ne 0, \tag 5$
forcing
$a + b = 0 \tag 6$
or 
$a = -b; \tag 7$
so 
$a = \pm b. \tag 8$
3:
If 
$a = a^{-1}, \tag 9$
then
$a^2 = 1, \tag{10}$
or
$(a - 1)(a + 1) = 0; \tag{11}$
now in an integral domain, if $a - 1 \ne 0$, (11) implies
$a = -1; \tag{12}$
so 
$a = \pm 1. \tag{13}$
4:
Finally, if $a \ne 0$, 
$a^n = 0 \Longrightarrow a(a^{n - 1}) = 0 \Longrightarrow a^{n - 1} = 0; \tag{14}$
a simple inductive argument now allows us to conclude that
$a = 0; \tag{15}$
i.e., if for some $k \in \Bbb N$,
$a^k = 0 \Longrightarrow a = 0, \tag{16}$
then
$a^{k + 1} = 0 \Longrightarrow a(a^k) = 0 \Longrightarrow a^k = 0 \Longrightarrow a = 0. \tag{17}$
A: Hint


*

*Note that $(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2.$ If $ab+ba=0$ then ...

*$a^2=b^2\implies b^2-a^2=0\implies (b-a)(b+a)=0\implies ...$ (use that the ring is an integral domain).

*From $2$ we have that $a^2=1\implies a=\pm 1.$

*$a^n=0\implies a=0$ (if $n=1$) and $a^n=0\implies aa^{n-1}=0$ (if $n\ge 2$) from where ... (use that the ring is an integral domain).

