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I am working on my scholarship practice exam which assumes high school or pre-university math knowledge and stuck at this question. Could you please have a look?

Let $p=n^2-18n+77$ for a natural number $n$. If $p>0$ and $p$ is prime, then $p=.....$

I started by factorizing the equation to $(n-7)(n-11)$ and drew the graph below.

I am not sure how to continue here. It seems that when $p>0$ or above $x$-axis, $p$ can be many values which are prime, seeing from the graph.

The answer provided is $5$. Please advise.

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    $\begingroup$ Remember your definitions and what it means for a number to be prime. $\endgroup$ – JMoravitz Jun 17 at 13:46
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Hint: You know $p=(n-7)(n-11)$, so if $p$ is prime the two factors $n-7,n-11$ must be $1,p$ or $-1,-p$ in some order.

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    $\begingroup$ Yes, I got it now. By applying the definition of prime number that briefly is the natural number that can be formed by multiplying only $1$ and itself, one of the terms would therefore equal $1$ and another would equal $p$ and check if $p$ is prime. Otherwise, swap the order and try again. Thank you for your hint. And apologies the question might be too amateur. I am not majoring in maths and trying my best to self study it. Thank you again, it helps a lot :) $\endgroup$ – Trey Anupong Jun 17 at 13:59

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