Proof that if $(n+1)^2 -1$ is even then $n$ is even?

Question

The forward implication, if $n$ is even then $(n+1)^2 -1$ is even, was simple.

I can't figure out the other implication: if $(n+1)^2 -1$ is even then $n$ is even.

What type of proof do I want to approach this with and where would I start?

Answer 1

If $(n+1)^2-1$ is even, then $(n+1)^2$ is odd. But if the square of something is odd, then the something is odd. (Try to prove that) Hence $n+1$ is odd and $n$ is even.

Answer 2

$(n+1)^2-1 = n^2+2n+1-1=n^2+2n=n(n+2)$.

$n$ and $n+2$ are either both even or both odd. The product of two odd numbers returns an odd number, while the product of two even numbers returns an even number. Hence if $n$ is even, then $n(n+2)$ is even, then $(n+1)^2-1$ is even, and vice versa.

Answer 3

If $n$ is odd, $n+1$ is even, which means $(n+1)^2$ is even, which means $(n+1)^2-1$ is odd. Now take the contrapositive of this statement.

Answer 4

Using the difference of squares factorization method we have: $$(n+1)^2-1=(n+1-1)(n+1+1)=n(n+2)=n^2+2n$$ Since $(n+1)^2-1$ is even, we can say $(n+1)^2-1=2a$ for suitable $a\in\Bbb N$. So we obtain $$n^2+2n=2a \Rightarrow n^2=2(a-n)$$ Thus, we see that $n^2$ is an even number (it is divisible by $2$), and so $n$ must also be even.

Answer 5

You want to prove the statement $[(n+1)^2-1\equiv0\pmod2]\implies[n\equiv0\pmod2]$.

Instead, prove the equivalent statement $[n\not\equiv0\pmod2]\implies[(n+1)^2-1\not\equiv0\pmod2]$:

$n\not\equiv0\pmod2\implies$

$n+1\equiv0\pmod2\implies$

$(n+1)^2\equiv0\pmod2\implies$

$(n+1)^2-1\not\equiv0\pmod2$