Question

Use mathematical induction to prove that the statement is true for every positive integer n. Show your work.

2 is a factor of n2 - n + 2

Answer 1

I will define: \(\color{#000000}{ \displaystyle f(n)=n^2-n=n(n-1) }\)

Write even positive integers in a form of \(\color{#000000}{ \displaystyle 2k}\),
and odd positive integers in a form of \(\color{#000000}{ \displaystyle 2k+1}\).\(\\[0.5em]\)
(where \(\color{#000000}{ \displaystyle k}\) is also some/any positive integer)
\(\\[0.9em]\)
When you plug \(\color{#000000}{ \displaystyle 2k}\), you get:
\(\color{#000000}{ \displaystyle f(2k)=2k(2k-1) }\)

When you plug \(\color{#000000}{ \displaystyle 2k+1}\), you get:
\(\color{#000000}{ \displaystyle f(2k+1)=2k(2k+1) }\)

So, regardless of whether \(n\) is odd or even,
when you plug in (any) \(n\), your output will
be always divisible by 2. That's because, for
any \(n\) - whether even or odd, your expression
\(f(n)\) is always a product of an even×odd
(where \(2k\) is an even).