The point of the final step is to prove that, upon squaring, $\rm\:b\:$ and $\rm\: 10-b\:$ leave the same remainder modulo $10\:,\:$ i.e. $\rm\:(\pm b)^2 = b^2\pmod{10}\:.\:$ Indeed $\rm\:(10-b)^2 =\: b^2 + 10\ (10-2\:b)\:.\:$ Therefore to find all of the squares modulo $10\:,\:$ it suffices to square only the integers between $0$ and $5\:,\:$ i.e.
$$\:\{0,1,2,3,4,5,6,7,8,9\}^2\equiv\{0,\pm1,\pm2,\pm3,\pm4,5\}^2 \equiv \{0,1,4,9,6,5\}\pmod{10}$$
since $\rm\:\ 9\equiv -1,\:\ 8\equiv -2,\:\ 7\equiv -3,\:\ 6\equiv -4\pmod{10}\:.$
The idea behind a proof by cases is to prove a statement $R$ seeing that, under the hypothesis, there are two possible cases (argument is similar for more cases); where the first case is the statement $P$, and $Q$ is the second. Thus, if we prove $P \implies R$ and $Q \implies R$, then we can conclude that $R$ is necessarily true. We can summarize with
$$(P \implies R) \land (Q \implies R) \quad\implies\quad (P \lor Q) \implies R\,.$$
Since $T \lor \neg T$ is a tautology, it is common to use this to take tha possible cases.
I can show you some examples of proof by cases in set-theory and real-analysis, but I do not if this is what you are looking for.
Remember, the key to prove by cases $P \implies Q$ is to split $P$ in all possibles cases such that
$$P_1 \lor P_2 \lor P_3 \lor \dotsm \lor P_n \iff P,$$
then prove $P_i \implies P$ where $i = 1, 2, 3, \dotso, n$.
In your problem, we need to use our "advanced knowledge" to pick the cases for $x^2 \ne 5$. Thus, since we know that for any numbers $a$ and $b$, only one of the three statements $a < b$, $a = b$ or $a > b$ is true. Since we have $x^2 \ne 5$, then must be either $x^2 < 5$ or $x^2 > 5$, but no both. For $x^2 < 5$, we have $S := \{x \in \Bbb Z: -2 \le x \le 2 \}$ satisfies the statement. For $x^2 > 5$, we have $T := \{x \in \Bbb Z: -3 \le x \text{ and } x \ge 3 \} = \{x \in \Bbb Z: -2 < x \text{ and } x > 2 \} $ satisfies $x^2 > 5$. Since $S \cup T = \Bbb Z$, we have our cases, i.e., we have $x \in S \implies x^2 < 5$ and $x \in T \implies x^2 > 5$, which is the same that $(x \in S \text{ or } x \in T) \implies (x^2 < 5 \text{ and } x^2 > 5)$, which is the same that $x \in \Bbb Z \implies x^2 \ne 5$.
Obviusly, this is not a rigorous proof. Also, you can to use the absolute value to ease the notation.
Best Answer
try this :-
first see what is $n^4$ for first integers $n$
$ 0^4 ={\color{Red} 0}\\ 1^4 ={\color{Red} 1}\\ 2^4=1{\color{Red} 6}\\ 3^4=8{\color{Red} 1}\\ 4^4=25{\color{Red} 6}\\ 5^4=62{\color{Red} 5}\\ 6^4=129{\color{Red} 6}\\ 7^4=240{\color{Red} 1}\\ 8^4=409{\color{Red} 6}\\ 9^4=656{\color{Red} 1}\\ 10^4=1000{\color{Red} 0}\\ 11^4=1464{\color{Red} 1}\\ 12^4=2073{\color{Red} 6}\\ ...$
Proof :- let $n_i$ be any integer number s.t $i$ is number of digits then $n_i=a_{i-1} 10^{i-1}+a_{i-2}10^{i-2}+...+a_110^1+a_010^0$ ,s.t $a_j$ is unit digit
then $n_i \bmod 10=a_{i-1} 10^{i-1}+a_{i-2}10^{i-2}+...+a_110^1+a_010^0 \bmod 10= {\color{Red} {a_0}} \bmod 10$ thus :- $n_i^4\bmod 10 ={\color{Red} {a_0^4}} \bmod 10 $
but $ 0\le a_0 \le 9 $ so we would have 10 cases for $n_i^4\bmod 10$
by cases :-
$ n_i^4\bmod 10 = 0^4 \mod10 ={\color{Red} 0}\mod 10\\n_i^4\bmod 10 =1^4 \mod 10 ={\color{Red} 1}\mod 10\\ n_i^4\bmod 10 =2^4 \bmod 10= {\color{Red} 6}\bmod 10\\ n_i^4\bmod 10 =3^4\bmod 10= {\color{Red} 1}\bmod 10\\ n_i^4\bmod 10 =4^4\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^4\bmod 10 =5^4\bmod 10= {\color{Red} 5}\bmod 10\\ n_i^4\bmod 10 =6^4\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^4\bmod 10 =7^4\bmod 10= {\color{Red} 1}\bmod 10\\ n_i^4\bmod 10 =8^4\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^4\bmod 10 =9^4\bmod 10= {\color{Red} 1}\bmod 10\\$
other method to prove it , $n_i^4=(n_i^2)^2$ and last digit of a square $\in \left \{ 0,1,4,9,5,6 \right \} $ so you would have 6 cases to check
$n_i^2\bmod 10 =0^2 \bmod 10= {\color{Red} 0}\bmod 10\\ n_i^2\bmod 10 =1^2\bmod 10= {\color{Red} 1}\bmod 10\\ n_i^2\bmod 10 =4^2\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^2\bmod 10 =5^2\bmod 10= {\color{Red} 5}\bmod 10\\ n_i^2\bmod 10 =6^2\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^2\bmod 10 =9^2\bmod 10= {\color{Red} 1}\bmod 10\\$