Division residue proof

If n is a natural number that gives a division of 2 or 3 when divided by 5, then n gives a residue of 4 when divided by 5.
Prove directly

Final Answer:

x =  1

Step-by-step explanation:

$$\begin{gathered} n\in N \\ n=5\cdot k_1+2\vee 5\cdot k_2+3\Rightarrow n^2=5\cdot k_3+4 \\ {k}_1,k_2,k_3\in N \\ (5\cdot k_1+2{)}^2=25\cdot k_1^2+20\cdot k_1+4=5\cdot (5\cdot k_1^1+4\cdot k_1)+4=5\cdot k_{31}+4 \\ {k}_{31}=5\cdot k_1^1+4\cdot k_1 \\ (5\cdot k_2+3{)}^2=25\cdot k_2^2+30\cdot k_2+9=25\cdot k_2^2+30\cdot k_2+5+4=5\cdot (5\cdot k_2^2+6\cdot k_2+1)+4=5\cdot k_{32}+4 \\ {k}_{32}=5\cdot k_2^2+6\cdot k_2+1 \\ x=1 \end{gathered}$$

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