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Prove or disprove : There is no positive integer $\ a\ $, such that $\ a^4+1\ $ is a Carmichael number.

Since $\ 3\ $ is not a weak Fermat-psedudoprime of $\ a^4+1\ $ upto at least $\ a=5\cdot 10^7\ $ (which I tested with pari/gp using the strict primality test, for "ispseudoprime" , which is a very reliable test, I arrived at $\ a=3\cdot 10^8\ $) , a Carmichael number is not possible until this limit.

I also tried to use Korselt's criterion which in this case menas that every prime factor $\ q\ $ of $\ a^4+1\ $ must satisfy $\ q-1\mid a^4\ $ , but this lead to nowhere.

Peter
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  • All these kinds of problems collide at the border of the unlimited (infinite), unless the numbers of the form $x^4+1$ are fully characterized, which is itself a problem of this kind. – Ataulfo Aug 30 '19 at 17:28
  • @Piquito There might be a trick that I overlooked. Moreover, perhaps there is a counterexample wihtin reach of computation. And finally, perhaps there is a property of Carmichael numbers that we can use here. So the situation need not be hopeless. – Peter Aug 31 '19 at 09:58
  • A possible key could be an exhaustive definition of the binomial $x^4+1$, maybe with some theory of algebraic numbers. It is known that the quadratic field $\mathbb Q (\sqrt2)$ is of unique factorization and perhaps the equality $(x^2+\sqrt2 x+1) (x^2-\sqrt2 x+1) = x^4 +1$ could be used. It is also known that an integer of this form has to be a product of squares and sums of squares as all sum of two squares but it is a particular one of sum of two fourth powers which is more difficult to characterize. – Ataulfo Sep 01 '19 at 14:13

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