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Suppose $k$ is a positive integer and $N:=k^4+1$ is composite.

Can $N$ be a $3$-Fermat-pseudoprime; that is, can the congruence $$3^{N-1}\equiv 1 \mod{N}$$ hold ?

I checked up to $k=10^8$ and found no example. Is this simple test in fact sufficient for numbers of the form $k^4+1$ ?

mjqxxxx
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Peter
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  • Interesting.. if I'm reading correctly, $3^{N-1}\equiv 1 \mod N \Rightarrow N$ is prime – AgentS Aug 28 '19 at 08:08
  • @ganeshie8 In fact, I hope this is the case for those numbers. – Peter Aug 28 '19 at 11:36
  • I extended the search limit to $3\cdot 10^8$, assuming we can trust the ispseudoprime-command for the numbers $N$ in this range. – Peter Aug 29 '19 at 20:43
  • @Peter I've added [examples-counterexamples] tag to few of your questions, feel free to attach it elsewhere you find suitable. – Sil Feb 01 '20 at 13:20
  • Search range extended to $k=2\cdot 10^9$ – Peter Jan 19 '23 at 10:36
  • @AgentS this is not what Fermat's Little Theorem means. It means if $N$ is a prime, then $3^{N-1} \equiv 1 \mod N$ and not necessarily the other way round. See http://oeis.org/A005935 for examples of composite numbers that satisfy the relation (couldn't find $k^4+1$ forms). – D S Jan 19 '23 at 13:54
  • @DS This comment refers to the form of the number in the question : $N=k^4+1$ with positive integer $k$ – Peter Jan 19 '23 at 13:56
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    @Peter The comment makes a common misconception that $a^{N-1} \equiv 1 \mod N \implies N$ is a prime. Just pointing out that. – D S Jan 19 '23 at 13:58
  • Due to some congruences we get restrictions on $k$, e.g. take 10, we have that $3^{10m-1} \bmod {10m}$ will always have a 3 or a 7 as ending digit and the congruence can thus never hold if $10\mid N$, for 34 this means that $k \bmod 34$ cannot be in ${9,15,19,25}$, 82 gives $k \bmod 82 \notin {3,27,55,79}$, $k \bmod 146 \notin {51,63,83,95}$, $k \bmod 386 \notin {9,43,343,377}$, and similar congruences cannot hold with 1681 and 94177 as modulo (and possibly others). This sadly only reduces the amount of $k$ to check by a little less than 20%. For $N=k^3+1$ this is over 68%. – jorisperrenet Jan 19 '23 at 20:26
  • In fact, 178, 194, 226, 274, 289, ... are other such numbers, this way 33% of all $k$'s can be left unchecked (note that it also takes time to do these modulo operations, thus too many would slow down the algorithm). – jorisperrenet Jan 19 '23 at 21:59
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    Checked all $k < 2 \cdot 10^{11}$. – jorisperrenet Jan 20 '23 at 10:01
  • @jorisperrenet A huge search limit. Apparently , my computer is too old for such calculations. I needed already quite long for the humble $2\cdot 10^9$. Anyway, thanks. – Peter Jan 20 '23 at 10:03

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