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Given the odd prime numbers,

Prove that if $x$ and $y$ are adjacent odd primes in this list, then $x + y$ has $3$ prime factors. The factors need not be distinct.

Here is an example I have provided: $3 + 5 = 8 = 2 \cdot 2 \cdot 2$. Therefore, $8$ has $3$ repeated factors of $2$.

Alexis Dailey
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  • By odd adjacent primes do you mean the neighbouring odd primes? .... Yeah actually I think you do, forget that I asked. – Mr Pie Feb 18 '18 at 04:58

3 Answers3

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If $x$ and $y$ were consecutive odd primes then $x+y = 2z$ for some integer $z$. If $z$ was a prime that would make $z$ lie between $x$ and $y$ which is not possible since $x,y$ are consecutive primes. Hence, $z$ has atleast two prime factors (not necessarily distinct).

sciona
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    I think it would be much better not to answer questions at this level that show no effort. – WillO Jan 14 '15 at 01:26
  • @WillO That is what you think. Not every body thinks that asking for showing effort is not an stupid idea. – Pp.. Jan 14 '15 at 01:27
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    @WillO isn't it a better idea to encourage the user asking the Q to show efforts first than discourage users from answering them ? :-) – sciona Jan 14 '15 at 01:29
  • Yeah, the show-effort army is "that rational". – Pp.. Jan 14 '15 at 01:31
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    Sciona: I think that discouraging users from answering them is prerequisite to encouraging effort. – WillO Jan 14 '15 at 01:31
  • @WillO: Thanks, I misread adjacent as twin. – André Nicolas Jan 14 '15 at 01:33
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    @Sciona: Plus, this seems to be a particularly egregious case since the comment on Meelo's answer indicates that the OP hasn't stopped to think about this question for an instant. – WillO Jan 14 '15 at 01:33
  • @WillO It is not your job. You are not anybody's daddy. It is their own decision the amount of effort each one want's to put on a problem. – Pp.. Jan 14 '15 at 01:35
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    @Pp.. Yes. and it's my own decision how I want to respond to that. – WillO Jan 14 '15 at 01:44
  • @WillO Andre's comment 'Multiplicity 4.' is ambiguous. When we talk about multiplicity of a prime dividing $p$ dividing $n$, we are looking for the highest power of $p$ in prime factorization of $n$. He possibly meant number of prime factors is $4$. – sciona Jan 14 '15 at 01:58
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$$11+13=24=2^3\cdot 3\,\,\,\,$$

Milo Brandt
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Suppose $x<y$ are adjacent odd primes. Then $x+y = 2a$, $a\in\mathbb{Z}$.

$x<y \Rightarrow 2x<x+y<2y \Rightarrow 2x<2a<2z \Rightarrow x<a<y$

How $x$ and $y$ are adjacent primes, then $a$ is not prime. Suppose $a=b\cdot c$, with $b, c>1$.

So $x+y = 2\cdot b\cdot c$. Therefore $x+y$ has at least 3 prime factors

Tiago Emilio Siller
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