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Let $P$ be a set of primes, such that for each nonnegative integer $n$, $19⋅8^n+17$ is divisible by some prime $p$ in $P$. Find the smallest possible number of elements in $P$. How do I start the problem? Thanks.

mathgeek
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  • @Bill Dubuque: I've read the other question several times, and I don't think that it's a duplicate of this question. In fact, except for the expression at hand - $19\cdot8^n+17$ - which is indeed identical, I don't quite see the "duplicate-ness". This question asks to find the smallest set of primes that can divide the expression for any natural value of $n$. The other question asks to prove that the expression is not prime for any natural value of $n$. So except for the expression itself, along with the relation to prime numbers, I think that it can hardly be considered duplicate. – barak manos Apr 07 '15 at 14:11
  • @Bill Dubuque: Continuing the previous (long) comment, can you please reconsider this? P.S., I admit that I might be "un-objective" here due to my answer below, but I honestly think that the two questions are not the same. Thanks. – barak manos Apr 07 '15 at 14:15
  • @barak The covering congruences arguments easily yield the result. But I'll undo my vote per your request. – Bill Dubuque Apr 07 '15 at 14:26
  • See this answer to a very closely related question for the standard technique of covering congruences (see esp. the paper of Schinzel linked there). – Bill Dubuque Apr 07 '15 at 14:31
  • @BillDubuque: Thanks (for both). – barak manos Apr 07 '15 at 14:38

1 Answers1

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The smallest possible number of elements in $P$ is $3$, with $P=\{3,5,13\}$:

  • $n\equiv0\pmod4\implies 3|19\cdot8^n+17$
  • $n\equiv1\pmod4\implies13|19\cdot8^n+17$
  • $n\equiv2\pmod4\implies 3|19\cdot8^n+17$
  • $n\equiv3\pmod4\implies 5|19\cdot8^n+17$

Let's prove each one of these statements by induction.

First, show that this is true for $n=0$:

  • $\small{19\cdot8^{0}+17= 3\cdot 12}$
  • $\small{19\cdot8^{1}+17=13\cdot 13}$
  • $\small{19\cdot8^{2}+17= 3\cdot 411}$
  • $\small{19\cdot8^{3}+17= 5\cdot1949}$

Second, assume that this is true for $n$:

  • $\small{19\cdot8^{4n+0}+17= 3a}$
  • $\small{19\cdot8^{4n+1}+17=13b}$
  • $\small{19\cdot8^{4n+2}+17= 3c}$
  • $\small{19\cdot8^{4n+3}+17= 5d}$

Third, prove that this is true for $n+1$:

  • $\small{19\cdot8^{4n+4}+17=4096\cdot(\color{red}{19\cdot8^{4n+0}+17})-69615=4096\cdot\color{red}{ 3a}-69615= 3\cdot(4096a-23205)}$
  • $\small{19\cdot8^{4n+5}+17=4096\cdot(\color{red}{19\cdot8^{4n+1}+17})-69615=4096\cdot\color{red}{13b}-69615=13\cdot(4096b- 5355)}$
  • $\small{19\cdot8^{4n+6}+17=4096\cdot(\color{red}{19\cdot8^{4n+2}+17})-69615=4096\cdot\color{red}{ 3c}-69615= 3\cdot(4096c-23205)}$
  • $\small{19\cdot8^{4n+7}+17=4096\cdot(\color{red}{19\cdot8^{4n+3}+17})-69615=4096\cdot\color{red}{ 5d}-69615= 5\cdot(4096d-13923)}$

The smallest possible number of elements in $P$ is $3$ because:

  • The base-cases show that there must be at least $3$ elements in $P$
  • The induction-steps show that the elements $3,5,13$ are sufficient
barak manos
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