proof verification - Divisibility of a statement without the induction principle

I want to prove that the following statement is divisible by $4$ with a direct proof.

$1 + (-1)^n ( 2 n -1 )$, $n$ natural number.

My solution : Because $n$ is a natural number, we can look at two individual cases: one where $n$ is an odd number, and one where $n$ is even. If both of these cases are divisible by $4$, then the aforementioned is divisible by $4$ also.

1 : If $n$ is odd, then the statement can be reduced to $2 ( 1 - n ).$ This is divisible by $4$ if it is divisible by $2$ twice. Divide by $2$ and we have $1 - n = 1 + ( - n ).$ The sum of two odd numbers is even and all even numbers are divisible by $2$. Therefore the statement is divisible by $4$ when $n$ is odd.

2 : If $n$ is even, the statement can be reduced to $2n.$ As in part $1,$ we show that this is divisible by $$ twice. $\frac{2n}{2} = n .$ Now since $n$ is an even number, it is divisible by $2$ and the statement is thus divisible by $4.$

Since 1 and 2 are both correct the statement is divisible by $4$.

Is this solution correct?

Answer

Your solution is correct. Here is an alternative (streamlined) wording.

To prove that the expression $1+(-1)^n(2n-1)$ is divisible by $4$ for all natural numbers $n$,

note that $n$ is either odd or even.

If $n$ is odd, then the expression is $2-2n=2(1-n)$, and $1-n$ is even,

so the expression is divisible by $4$.

If $n$ is even, then the expression is $2n$, so it is divisible by $4$.

In any case, the expression is divisible by $4$. QED