Divisibility of a statement without the induction principle

Question

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 1

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

Answer 2

Yes, it is correct. Another way: $\bmod \color{#c00}4\!:\ (-1)^n\equiv 1\! -\! 2n\,$ by parity case analysis, so it boils down to $\, \underbrace{(1\!-\!2n)^2}_{\large \rm odd^{\large 2}} = 1\!+\!\color{#c00}4n(n\!-\!1)\equiv 1.\, $ In fact $\,\bbox[4px,border:1px solid #c00]{{\rm odd}^2\equiv 1\pmod{\!\color{#0a0}8}}\,$ by $\,2\mid n(n\!-\!1),\,$ a handy lemma.

Or we could add $\,4n\,$ so $\,(-1)^n\equiv 1\!+\!2n\,$ and $\, \underbrace{(1\!+\!2n)(1\!-\!2n)\equiv 1-4n^2}_{\bbox[4px,border:1px solid #c00]{\large \text{difference of squares}}}\equiv 1 \pmod{\!4}$

Both "squaring" ideas are ubiquitous in number theory so are worth knowing well.