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Question: Find all the prime numbers of the form $n^2 + 4n$. List of the primes of this form and prove these are all such primes.

My Answer I'm not really good at this but I made an attempt. $$n^2 + 4n = n(n+4)$$ since a prime number is divisible by 1 and itself, $$\frac{n(n+4)}{n} = \frac{n}{n}\cdot\frac{n+4}{n} = 4$$ Since $4$ is not a prime number, no prime numbers of the form $n^2 + 4n$ can be listed.

please guide, thanks.

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    $\begingroup$ Your first step is correct, factoring. But you missed $5$ as an answer, so back up. $\endgroup$ – GFauxPas Apr 13 '15 at 23:36
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If $n^2+4n=n(n+4)$ is a prime number then it's divisible by $n$ and $n+4$ but also only 1 and itself. The conclusion is $n=1 , n+4=5 \Rightarrow n^2+4n=5$ and that is a prime number. Only $5$ satisfies this.

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Since $n^{2} + 4n = n(n+4)$, $n$ and $n+4$ both divide your number. If you assume your number is prime, then $n$ must be $1$ and $n+4$ must be your number. thus the only prime you're looking for is $5$.

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$n^2 + 4n = n (n + 4)$ is the product of two numbers, n and n+4. The product of two integers is composite except when one of them is 0 or 1. n = 1 gives the prime number 1 * (1 + 4) = 5.

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