5
$\begingroup$

I am trying to write a proof for this theorem:

For every positive integer $n$, $n^2+4n+3$ is not a prime.

Proof: Let $n \in \mathbb{Z}^{+}$. Note that $$n^2+4n+3=(n+1)(n+3)>1\text{,}$$ and $n+1 >1$ and $n+3 >1$.

Let $a = n+1$ and $b = n+3$. Then we have $$\dfrac{(n+1)(n+3)}{a}>\dfrac{1}{a}$$ and $$\dfrac{(n+1)(n+3)}{b}>\dfrac{1}{b}\text{.}$$ Therefore, $n^2+4n+3$ is not prime. $\square$

I don't think my proof is right and miss many things. Can anyone give me a hit or show me how to write a better proof for this question?

$\endgroup$
2
  • $\begingroup$ I would think that, unless your proof has to be extremely rigorous, it might do just to show neither factor is $1$, nor $-1$. $\endgroup$
    – Mike
    Feb 11, 2015 at 7:05
  • 1
    $\begingroup$ @Simple When you wrote "Therefore $n^2+4n+3$ is not prime", what was that based on? A proof should try to connect with the definition of prime at some point, and yours does not. $\endgroup$
    – Erick Wong
    Feb 11, 2015 at 7:08

2 Answers 2

11
$\begingroup$

I don't know what your definition of prime is. According to Wikipedia, a prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself. You have already factored the expression as $(n+3)(n+1)$. It should suffice to note that since $n>0, n+3>n+1>1$. Since your expression has at least $2$ positive factors, neither of which is $1$, it follows that $(n+3)(n+1)$ is not prime.

$\endgroup$
1
  • 1
    $\begingroup$ @Simple I would consider this a much better answer than my own--much more concise. The way I did it is simply another approach and, as I mentioned, is often a good way of going about things when you are really stuck. Now you can see how your claim may be proved in a variety of ways. (+1) $\endgroup$ Feb 11, 2015 at 7:52
1
$\begingroup$

One easy way of going about this is simply to consider the parity of $n$:

$n$ is even: We have that $$ (2\ell)^2+4(2\ell)+3=4\ell^2+8\ell+3=\underbrace{(2\ell+1)(2\ell+3)}_{\text{composite}}. $$ Thus, $n^2+4n+3$ is not a prime when $n$ is even.

$n$ is odd: We have that $$ (2\ell+1)^2+4(2\ell+1)+3=\underbrace{2(2\ell^2+6\ell+4)}_{\text{composite}}. $$ Thus, $n^2+4n+3$ is not a prime when $n$ is odd.

Consequently, when $n$ is even or odd, we have that $n^2+4n+3$ is not a prime. $\Box$

$\endgroup$
5
  • $\begingroup$ Why the downvote? $\endgroup$ Feb 11, 2015 at 7:00
  • $\begingroup$ Wow, I never think about this way.+1 $\endgroup$
    – Simple
    Feb 11, 2015 at 7:02
  • 2
    $\begingroup$ @Simple Oftentimes when I am having trouble proving a property for all $n$ where $n\in\mathbb{Z}$, I consider the even and odd case when all hope is lost. Sometimes this can be useful when using an induction argument where you induct on even and odd numbers (when simply inducting on $n$ by itself is tricky). Useful little thing to know. $\endgroup$ Feb 11, 2015 at 7:04
  • 3
    $\begingroup$ If the OP cannot conclude that $(n+1)(n+3)$ is composite, why is it safer to conclude that $(2m+1)(2m+3)$ is composite? What is the benefit of splitting this into 2 separate cases? $\endgroup$
    – Mike
    Feb 11, 2015 at 7:17
  • $\begingroup$ @Mike I actually did not bother to address the OP's attempted proof. The OP asked, "Can anyone give me a hint or show me how to write a better proof for this question?" I would consider what I have provided as a better proof. $\endgroup$ Feb 11, 2015 at 7:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .