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I am trying to find the mistake in the following "proof", which shows that a Fermat-pseudoprime $n$ with respect to $a$ is also a strong pseudoprime with respect to $a$.

"Write $n-1=2^st$ with $t\in\mathbb{Z}$ odd and $s\in\mathbb{Z}_{\geq0}$. Since $n$ is Fermat-pseudoprime with respect to $a$, we have that $a^t+n\mathbb{Z}$ is a root of the polynomial $$X^{2^s}-1=(X^{2^{s-1}}+1)\dotsb(X^2+1)(X+1)(X-1)\in(\mathbb{Z}/n\mathbb{Z})[X].$$ Therefore $a^t+n\mathbb{Z}$ is a root of $X-1$ or a root of a polynomial of the form $X^{2^j}+1$ with $0\leq j<s$. We conclude that $n$ is a strong pseudoprime with respect to $a$."

I suppose that the mistake lies in the argument "$a^t+n\mathbb{Z}$ is a root of the polynomial [...]", since I traced the proof with $a=2$ and $n=341$ in order to reach a contradiction somewhere, and that part is the only one that doesn't seem logical at once. (341 is a Fermat-pseudoprime with respect to 2, but not a strong pseudoprime.) However, I don't really see why the mistake is located there (if it is actually located there).

bitsbit
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1 Answers1

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For composite $n$, "$a^t+n\mathbb{Z}$ is a root of a product of some polynomials" does not imply "$a^t+n\mathbb{Z}$ is a root of one of these polynomials". It may well happen that the product, with each term evaluated at $a^t+n\mathbb{Z}$, just represents a (nontrivial) factorization of $0$ in $\mathbb{Z}/n\mathbb{Z}$.

metamorphy
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  • e.g. $\bmod 8!:\ 3,$ is a root of $,(x-1)(x+1).\ $ In fact the quadratic $,x^2-1,$ has four roots $,x\equiv \pm1,\pm3,,$ more roots than its degree (which occurs in $,\Bbb Z/n = $ integers $!\bmod n \iff n,$ is compositie, see here ) – Bill Dubuque Jun 22 '19 at 15:26