Proving properties of rings

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.

$$(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+b^2$$ $$(a-b)^2=(a-b)(a-b)=a^2-ab-ba+b^2=a^2+b^2.$$ $$(ab)^2=abab=-baab=-ba^2b=a^2bb=a^2b^2$$ $$(a-b)(a+b)=a^2+ab-ba-b^2=a^2+2ab-b^2.$$ $$a^2+a^2=(a+a)(a+a)=a^2+a^2+a^2+a^2.$$ Thus, $2a^2=0$ and from here we can not get $a^2=0$ for all ring.

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

$a^{k + 1} = 0 \Longrightarrow a(a^k) = 0 \Longrightarrow a^k = 0 \Longrightarrow a = 0. \tag{17}$