Prove that for every positive integer $n$, $38^n+31$ is a composite number. for example $38+31=69$ is composite. $38^2+31=1475$ is also composite. I have tried modulo but it didn't work.
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Well,it is divisible by 3 for odd n. – rah4927 Apr 22 '14 at 16:30
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Someone has asked a similar question a few month ago. We don't know whether this is true or not. However, according to some comments there, $38^n + 31$ is composite for $n \le 220,000$. – achille hui Apr 22 '14 at 16:39
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@achille,is this question suitable for MO? – rah4927 Apr 22 '14 at 17:11
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@Satvik,pray tell us where you have got this problem from. – rah4927 Apr 22 '14 at 17:12
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1My friend had given me this problem. – Satvik Mashkaria Apr 22 '14 at 17:13
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@rah4927 No idea, I'm not a mathematician ;-p But the other question has been there for a few months with 60 up-votes. If one points that out and rephrase the question properly, it is possible MO will accept it. – achille hui Apr 22 '14 at 17:48
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I voted to close as duplicate - maybe this question would be better merged with the previous one though. – Mark Bennet Apr 22 '14 at 18:23
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For odd $n$, take modulo $3$. For even $n$, if $n \equiv 2 \pmod 4$, take modulo $5$. So only $n \equiv 0 \pmod 4$ case remains. (We can further break it up into the cases $12n+4,12n+8,12n$ with modulo $7$..)
Anyway, it is hard to get a conclusive answer out of these types of problems....
Sandeep Silwal
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As noted in a comment this case as already been asked and it hasn't been solved yet, look the other question for more details – Alessandro Codenotti Apr 22 '14 at 17:10