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How can man proof this Theorem step by step, specially when $$p_i\mid a?$$ This theorem is in book: Elementary number theory by Koshy on page 334

Let $$p_1, p_2, . . . , p_k$$ be any distinct primes, and let $a$ any positive integer, and $$L = {\rm lcm}(p_1 − 1, p_2 − 1, . . . , p_{k} −1).$$ Then $$a^{L+1} \equiv a\pmod{p_1 p_2 \cdots p_k}.$$ i can not understand the proof in the book, Can anyone help me?

enter image description here [excerpt added by Bill D.]

Bill Dubuque
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  • Where does Phythian enter? – lhf Feb 25 '21 at 11:05
  • Why cannot we say $a^{L+1}-a \equiv 0 \pmod{p_1\dots p_k}$ is true just because $p_i \mid a$? – VIVID Feb 25 '21 at 11:08
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    @lhf, it's probably a reference in Koshy's book to a paper on the subject written by Phythian. – Barry Cipra Feb 25 '21 at 11:14
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    @lhf It seems to refer to a 1970 note in Math. Gazette" by J.E. Phythian, Math. Dep., Univ. Dar es Salaam, Tanzania. It's a very strange reference for such an old well-known result. – Bill Dubuque Feb 25 '21 at 11:40
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    Which step in the proof do you not understand? – Bill Dubuque Feb 25 '21 at 11:51
  • @BillDubuque, great find, thanks! – lhf Feb 25 '21 at 11:57
  • i want to understand This: When $$p_i \mid a$$ then $$a^{L+1}≡a\hspace{3pt} mod\hspace{3pt}(p_1.p_2p_3. ...p_k)$$ – AAA Feb 25 '21 at 12:17
  • Koshy's proof is incorrect if some $p_i\mid a,,$ likely the source of your doubt. To fix it add the case that $,p_i\mid a,\Rightarrow, ,p_i\mid a^{L+1}-a =:N,,$ so this divisibilty always holds - both when $,p_i\mid a,,$ and when it does not, by the given proof via little Fermat; typo fix: $,(a^{\color{#c00}{p_i-1}})^{\ell/(p_i-1)}\equiv 1\pmod{p_i}.\ \ $ – Bill Dubuque Feb 25 '21 at 13:13
  • Since $N$ is divisible by the distinct primes $p_i$ it is also divisible by their product $P$ (provable either by FTA = existence & uniqueness of prime factorizations, or lcm = product for coprimes, or by iterating CCRT or Euclid's Lemma, etc). Thus $,P\mid a^{L+1}-a,\Rightarrow, a^{L+1}\equiv a\pmod{! P}\ \ $ – Bill Dubuque Feb 25 '21 at 13:14
  • That case is also not handled in Phythian's paper, so it too is incorrect. It is a trivial case, for sure, but it needs to be mentioned in order to obtain a complete, correct proof. – Bill Dubuque Feb 25 '21 at 13:21
  • But how we can from $$p_i\hspace{3pt} \mid( a^{L+1}-a)$$ go to and proof that $$p_1.p_2.p_3....p_k\hspace{3pt}\mid( a^{L+1}-a)$$ – AAA Feb 25 '21 at 13:41
  • That is explained in my second last comment, which mentions a few methods to prove that any integer divisible by distinct primes is also divisible by their product. Note: you need to start your messages with "@Bill" for me to be notified that you have replied. Else I may never see them. – Bill Dubuque Feb 25 '21 at 14:01

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