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Suppose that $p$ and $p^{2}+14$ are both prime numbers. Prove that $p$ must be equal to $3$

What i tried

By the definition of prime numbers,

$$p=k_{1}l_{1}$$ $$p^{2}+14=k_{2}l_{2}$$

where $$k_{1}=0$$ or $$l_{1}=0$$

and $$k_{2}=0$$ or $$l_{2}=0$$

I tried combining both equations together to get $$(k_{1}l_{1})^{2}+14=k_{2}l_{2}$$

Im unsure of how to continue from here. Could anyone provide me a hint that would lead me to an answer. (but not giving me an answer as i want to try it myself). Thanks

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  • $\begingroup$ Hint: if $p>3$ is a prime then $p^2$ is one more than a multiple of $3$. $\endgroup$ – lulu Sep 29 '15 at 13:48
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Hint: If $p$ is not $3$, $p^2 + 14$ must be divisible by $3$ - why?

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You have unnecessarily complicated the issue. Any prime other than $3$ is of the form $3n \pm 1$, so $p^2$ = 1 (mod 3) and $14 = -1$ (mod 3)

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