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I was reading a question in one of the previous pages, in searching for a proof I stumble across what seem like a contradiction. All I want is for someone to provide the missing link in my argument.

The question

Find the least $n$ for which $38^n+31$ is prime.

My attempt at a proof

If $38^n+31$ composite, then there exist at least a prime $p$ such that $p|38^n+31$. Now $\gcd(p,38)=1$, otherwise, $d=\gcd(p,38)=2$ or $19$ and $d|31$ a contradiction. Hence, by Fermat's Little Theorem; $38^{p-1} \equiv 1 \pmod p$, and for all positive integer $r$, $38^{r(p-1)} \equiv 1 \pmod p$.

Hence, $38^{r(p-1)}+31 \equiv 32 \pmod p$, but $38^{r(p-1)}+31 \equiv 0 \pmod p$, because it's composite. It follows that $32 \equiv 0 \pmod p$ i.e $p|32$, a contradiction, and hence, the above expression cannot be composite (but inputing real values for $n$ shows that it is indeed composite).

Klangen
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user145342
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  • You went astray in $38^rp-1\equiv1 \pmod p$. How did you arrived there? – chubakueno Apr 24 '14 at 06:31
  • What exactly are you doing to the equivalence $38^{p-1}\equiv1$ to get $38^{rp-1}\equiv1$? – 2'5 9'2 Apr 24 '14 at 06:38
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    It also seems to me that you have actually "concluded" that no prime $p$ divides $38^n+31$ at all. So you're not just "proving" that it is not composite, but also that it is not prime. So it's not anything? – 2'5 9'2 Apr 24 '14 at 06:41
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    Is the issue that by $38^{rp-1}$ you actually mean $38^{r(p-1)}$? – 2'5 9'2 Apr 24 '14 at 06:44
  • @alex, that's what i meant, just that i have typing problems – user145342 Apr 24 '14 at 06:49
  • http://math.stackexchange.com/questions/597234/least-prime-of-the-form-38n31 – lab bhattacharjee Apr 24 '14 at 07:16
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    I think you've assumed that $n=r(p-1)$ for some $r$, which need not be the case, in the third to last line. –  Apr 27 '14 at 21:44
  • Agree with Alex Jordan, Bryan and kingW3. You have only shown that $n$ cannot be a multiple of $p-1$. So for a given $p$ we need to use a different $n$, but is there some $n$ that does not work for any $p$ (non-trivially)? – Jyrki Lahtonen Jun 01 '14 at 08:42

3 Answers3

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It does not follow that $38^rp-1\equiv 1\pmod p$. In fact $38^rp-1\equiv -1\pmod p$. And how are you trying to get from $38^rp-1+31$ to $38^n+31$ anyway?

Hagen von Eitzen
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Clearly $p-1\not=1$ since expression is odd,so $p\geq3$ clearly the expression doesn't cover the whole $\mathbb{N}$,you would have to prove these cases also $$r(p-1)+i,i=1,2,3\cdots p-2$$

kingW3
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For $38^x+31$, it is useful to note that the small primes divide $38^x+1$, as $x$ is odd or double-odd. So, eg $3 \mid 38^x+1 $ for odd x. $5 \mid 3d^x + 1$ for $x=2 \pmod 4$. So after removing 3 and 5, one is left with $4 \mid x$.

Running through the output of factor for $38^{4x}+31$, gives some rather difficult numbers to factorise. for x=3, there are three divisors, 338431, 322249, and 83126873.

It's probably not an easy row to hoe here. There well may be primes, but these are not generally accessable. factor seems to indicate that if it were prime, then $4x\gt 176$

hola
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wendy.krieger
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