Respuesta :
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].
//
Answer:
See below.
Step-by-step explanation:
If a = 5 mod 8 and b = 3 mod 8
then ab = 5*3 mod 8 = 15 mod 8 = 7 mod 8.
ab + 1 = 8 mod 8 = 0 mod 8 so it is divisible by 8.
