Proof question

[Faded-Maximus]

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

I have let k represent any positve integer.
therefore k + 3 > 3.
(k+3)^2 - 4
k^2 + 6k + 9 - 4
k^2 + 6k + 5

I am stuck here and don't know how to proceed. If somebody could help that would be great.

 

[skeeter]

k^2 + 6k + 5 = (k + 5)(k + 1) ... looks composite to me

 

[Faded-Maximus]

Didn't even think to factor it again. Thanks. I am only having difficulty with one last question.

Prove that 25 is subtracted from the square of an odd integer greater than 5, the resulting number is always divisable by 8.

My solution:
Let k be any postive number
therefore 2k + 5 > 5
(2k + 5)^2 - 25
4k^2 + 20k + 25 - 25
4k^2 + 20k
4k(k+5)

Once again I do not know where to go from here..

 

[soroban]

Faded-Maximus!

Prove that 25 is subtracted from the square of an odd integer greater than 5,
the resulting number is always divisable by 8.

My solution:

Let \(\displaystyle k\) be any postive number.
Therefore: \(\displaystyle \,2k + 5 \:> \:5\)

\(\displaystyle (2k\,+\,5)^2\,-\,25\)
. . \(\displaystyle = \;4k^2\,+\,20k\,+\,25\,-\,25\)
. . \(\displaystyle = \;4k^2 + 20k\)
. . \(\displaystyle = \;4k(k+5)\)

You've shown that the number is divisible by 4.

We must show that there is another factor of 2 in there . . .

If \(\displaystyle k\) is even, then \(\displaystyle 4(\text{even})(k+5)\) is divisible by 8.

If \(\displaystyle k\) is odd, then \(\displaystyle k\,+\,5\) is even.
. . Hence: \(\displaystyle \,4k(\text{even})\) is divisible by 8.

There!

 

[Faded-Maximus]

Thanks alot guys!