Answer:
See proof below
Step-by-step explanation:
Let [tex]n=4k+3[/tex] for some integer k.
Multiply n by itself to get [tex]n^2=(4k+3)(4k+3)=[16k^2+12k+12k+9=16k^2+24k+9[/tex]
Now substract 1 in both sides of the equation, and factor 8 to get
[tex]n^2-1=16k^2+24k+8=8(2k^2+3k+1)=8m[/tex], if we define [tex]m=2k^2+3k+1[/tex].
Thus, [tex]n^2-1=8m[/tex] for some integer m, that is, [tex]8|n^2+1[/tex]
