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.