Prove that if a is equivalent to 5 mod (8) and b is equivalent to 3 mod (8), then 8 divides ab+1

Answer:Explanation contains the proof.Step-by-step explanation:[tex]a \equiv 5 (mod 8) \text{ means there is integer } k \text{ such that } a-5=8k[/tex].[tex]b \eqiv 3 (mod 8) \text{ means there is integer } m \text{ such that } b-3=8m[/tex].We want to show that [tex]8 \text{ divides } ab+1[/tex].  So we are asked to show that there exist integer [tex]n \text{ such that } 8n=ab+1 \text{ or 8n-1=ab[/tex]So what is [tex]ab[/tex]?[tex]a-5=8k \text{ gives us } a=8k+5[/tex].[tex]b-5=8m \text{ gives us } b=8m+5[/tex].So back to [tex]ab[/tex]....[tex]ab[/tex][tex]=(8k+5)(8m+5)[/tex][tex]=64km+40k+40m+25[/tex]  (I use foil to get this)Factoring out 8 gives us:[tex]=8(8km+5k+5m)+25[/tex]Now I could have factored some 8's out of 25.  There are actually three 8's in 25 with a remainder of 1. [tex]=8(8km+5k+5m+3)+1[/tex]We have shown that there is integer [tex]n \text{ such that } ab=8n-1[/tex].The integer I found that is n is 8km+5k+5m+3.Therefore [tex]8|(ab+1)[/tex].//