Question

Show that n^2-1 is divisible by 8,if 'n' is an odd positive integer.

Answer 1

$n-an\ odd\ positive\ integer\ \ \ \Rightarrow\ \ \ n=2k+1\ \ \ and\ \ \ k\in Natural\\\\n^2-1=(2k+1)^2-1=(2k)^2+2\cdot2k\cdot1+1^2-1=4k^2+4k=\\\\=4k(k+1)\\\\if\ k-odd\ \ then\ (k+1)-even,\ \ \ then\ k(k+1)-even\\\\\ \ \ \Rightarrow\ \ \ k(k+1)=2m\ \ \ and\ \ \ m\in N\ \ \ \Rightarrow\ \ \ 4k(k+1)=4\cdot2m=8m\\\\\\if \ k-even,\ \ \ then\ k(k+1)-even\\\\\ \ \Rightarrow\ \ \ k(k+1)=2p\ \ \ and\ \ \ p\in N\ \ \ \Rightarrow\ \ \ 4k(k+1)=4\cdot2p=8p$

$------------------------\\\\ for\ each\ k \in N\ the\ product\ of\ \ \ k (k +1)\ \ \ is\ even,\ so:\\\\if\ 'n'\ is\ an \ odd\ positive\ integer,\ then\ \ \ n^2-1\ \ \ is\ divisible\ 8.$