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Background: This is from a proof that 341 is the lowest pseudoprime, meaning it passes the base test $a^n\equiv a \pmod n$ but n is not actually a real prime because $11\times 31=341$.

The author states:

(In fact, knowing this factorisation in advance, one could 'cheat' in the base 2 test to avoid large computations: since 11 and 31 are primes, Theorem 4.3 gives $2^{10} = 1 \mod (11)$ and $2^{30} = 1 \mod (31)$, which easily imply that $2^{341} - 2$ is divisible by both 11 and 31, and hence by 341.)

Was this obvious implication the chinese remainder theorem, or manually plugging one equation into the other, or is there some obvious rule that I don't know about that says $a_1\equiv b \mod n_1$ and $a_2\equiv b \mod n_2$ $\implies$ $a_1a_2\equiv b \mod n_1n_2$

Please don't try to prove the number is pseudoprime, I see an easy implication there ( any a will pass the test mod 11: $a^{10*34+1}\mod 11$, and I can simpify mod 31. I want to know if there's some trick that makes the first two congruences imply the third just but looking at them.

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    I think the author made a mistake. If instead we note that $2^5\equiv 1\bmod 31$, then it does indeed follow easily. – 2017-02-02
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    7= 2 mod 5, and 2=2 mod 13 but 14 != 2 mod 26. So you don't know that rule for a good reason. – 2017-02-02

3 Answers 3

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As $2^5\equiv1\pmod{31}\implies2^{10}=(2^5)^2\equiv1^2=1$

and$2^5\equiv-1\pmod{11}\implies2^{10}\equiv(-1)^2\equiv1$

So, $2^{10}-1$ is divisible by $11,31$ hence by their LCM.

As $n=2^{340}-1=(2^{10})^{34}-1^{34}$ is divisible by $2^{10}-1,2n$ will be divisible by LCM$(11,31)$

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    But this follows from $2^5\equiv 1 \bmod 31$, not from $2^{30}\equiv 1 \bmod 31$. – 2017-02-02
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    @TonyK, Exactly, That's why the Question says: "one could 'cheat'" – 2017-02-02
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    I don't follow you. – 2017-02-02
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    I don't know if this is what the authors had in mind, but this is the way to do it I think. Maybe they made a mistake (I haven't seen many yet in this particular book.) – 2017-02-02
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Hints:

Because $\;341=31\cdot11\;$ ...and thus

$$2^{341}-2=(2^{31})^{11}-2\ldots$$

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    yes but 2 is not a square, so you can't factor out $(2^{31}-1)$ – 2017-02-02
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    @use I know $\;2\;$ is not a square (but it is modulo $\;17\;$ , for example...), yet I didn't mean that, but: $$(2^{31})^{11}-2=(2^{31})^{11}-2^{11}=(2^{31}-2)((2^{31})^{10}-2\cdot(2^{31})^{9}+\ldots)$$ i.e. using the development from geometric progressions $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+b^{n-1})$$ – 2017-02-02
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    I don't think as fast as some of you answer. I'm still thinking about the other answer... – 2017-02-02
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    @user135711 I didn't say it is "as fast as...". It is a hint, but it doesn't matter: you already got other answers. :) – 2017-02-02
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    Ahhh! Also a good hint/"way to make it easily imply" – 2017-02-02
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$2^{10}\equiv 1 \mod 11$

So $2^{11}\equiv 2 \mod 11$

And $2^{31}\equiv 2 \mod 31$.

So $2^{341}=(2^{11})^{31} \equiv 2^{31} \equiv 2\mod 11$.

So $11|2^{341}-2$.

Likewise for 31.

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    This just proves $a \equiv b \mod n_1$ and $a \equiv b \mod n_2$ $\implies a\equiv b \mod n_1 n_2$. I.e. proving that the number is a pseudo-prime manually using the established rules. Although I do appreciate the answer very much. The authors made it seem like when $a_1$ and $a_2$ weren't equal you could multiply them or combine them. – 2017-02-02
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    Um, then I don't understand what your question was. – 2017-02-02
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    Well, as far as I've gotten in this subject, you can only combine primes in the mod base n, when all the a's are equal and all the b's are equal in $a \equiv b \mod n_i$ I was wondering if there was a new rule that I don't know about to combine $a_i\equiv b \mod n_i$ into $a_1a_2\dots\equiv b \mod n_1 n_2\dots$ that I haven't learned yet like it seemed was implied. My question might have been not very well written, but the text so far has been very precise about wording things, so I was thinking there was something I might be missing. – 2017-02-02
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    No, there is no such rule and the author wasn't implying any such rule. The author was simply say 2^10 = 1 mod 11, and 2^30 =1 mod 31 implies 2^341 - 2 = 0 mod 341. He concluded that by the reason I gave. No other. Your rule is obviously false – 2017-02-02
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    Yes, I didn't think that rule existed. Did you down vote my question? – 2017-02-02
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    No. I wouldn't do that. – 2017-02-02