Proving

[loser]

Prove that if 4 is subtracted from the square of an interger greater than 3, the result is a composite number.

I need to prove it deductively.

I know that an example could be written as:

(n+4)^2-4 = composite number

it is n+4 as the number has to be greater than 3 so if I can prove 4 first then i think i should be able to prove the rest.... let me know if i am on the right track.

 

[soroban]

Hello, loser!

Prove that if 4 is subtracted from the square of an interger greater than 3,
the result is a composite number.

Let $\displaystyle n$ be the integer greater than 3.
$\displaystyle \;\;$Then we examine: $\displaystyle \,n^2\,-\,4$

If $\displaystyle n$ is even, it is of the form: $\displaystyle \,n\:=\:2k$ for some integer $\displaystyle k\,\geq\,2$
$\displaystyle \;\;$then we have: $\displaystyle \,(2k)^2\,-\,4\:=\:4k^2\,-\,4\:=\:4(k^2\,-\,1)$ . . . a composite.

If $\displaystyle n$ is odd, it is of the form: $\displaystyle \,n\:=\:2k\,+\,1$ for some integer $\displaystyle k\,\geq\,2$
$\displaystyle \;\;$then we have: $\displaystyle \,(2k\,+\,1)^2\,-\,4\:=\:4k^2\,+\,4k\,+\,1\,-\,4$
\(\displaystyle \;\;\;\;=\:4k^2\,+\,4x\,-\,3\:=\2k\,-\,3)(2k\,+\,1)\) . . . a composite.

 

[Matt]

There's no need to consider two cases. Just write:

\(\displaystyle \L\qquad n^2-4=(n+2)(n-2),\)

which is clearly composite if n is larger than 3.