Well that's the question I am trying to solve. I did check it for a few $q$ and it seems to hold. However, I'm not sure how I would go about proving this. I actually cannot figure out where to start. I tried adding and subtracting $2q$ to make a perfect square. I think I might have to use mod 10 in this to make the algebra a little easier but other than that I don't know where to start. Any hints/tips/suggestions?
For $q \geq 3$, if $q^2 + 1$ is prime then the final digit of $q$ is $0,4,$ or $6$ for $q \in \mathbb{N}$
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number-theory
prime-numbers
2 Answers
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First of all, if $q$ is odd, then $q^2+1$ is ... and therefore can't be prime. Thus we only need to consider even values of $q$. All even numbers end with $0$, $2$, $4$, $6$, or $8$. And apparently, we only need to explain why the last digit can't be $2$ or $8$. Hint: if the last digit of $q$ is $2$ or $8$, what is the last digit of $q^2+1$?
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0It would 5 right? Which means q isn't prime because it's not odd either. – 2017-01-26
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0@Chubbles: Yes, it would be $5$. No, it doesn't mean that. $q$ isn't supposed to be prime, $q^2+1$ is. If $q^2+1$ ends with a $5$, then it's divisible by ... and thus cannot be prime. (Unless it is the single-digit number $5$ itself. But $q\ge3$ implies $q^2+1\ge10$, so we're safe.) – 2017-01-26
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${\bf Hint}\quad 10 So $q$ has $\color{#90f}{\rm even}$ units digit, and $\,\color{#0a0}{q\not\equiv 2,3\pmod{\!5}} = \color{#c00}2,3,7,\color{#c00}8\pmod{\!10}\,$ excludes even digits $\,\color{#c00}{2,8}$ Remark $\ $ Note the we didn't need primality but, rather, coprimality to $10$