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Suppose that we have $a=p_1 \cdot p_2\cdot p_3=q_1\cdot q_2\cdots q_s$ where $p_1,p_2,p_3$ and $q_1,q_2,...,q_s$ are primes. Explain why $s=3$.

This seems like a pointless question, but is it $3$ because $p_n$ doesn't go past $p_3$?


I'm still having a hard time with this question. I'm mainly not understanding what it is asking.

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  • $\begingroup$ Please check to see that I didn't alter the intended meaning with my edit. $\endgroup$ – Cameron Buie Dec 10 '12 at 2:10
  • $\begingroup$ Yes, there was an edit that changed the meaning. $\endgroup$ – Dmitri.Mendeleev Dec 10 '12 at 2:13
  • $\begingroup$ Aha! Now it makes sense. $\endgroup$ – MJD Dec 10 '12 at 2:14
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I don't think this question is posed correctly, because it's definitely not true: $$30=2\cdot3\cdot5=13+17\Rightarrow s=2$$ and $$42=2\cdot3\cdot7=5+7+13+17\Rightarrow s=4$$ are just two counterexamples.

For the corrected post: We want to show that if $a=p_1p_2p_3=q_1q_2\dots q_s$, then $s=3$. Clearly, $a$ divides $p_1$, so by the theorem on prime division, one of the $q_i$ divides $p_1$. Since $q_i$ is prime, then, $q_i=p_1$ and we can cancel it to get $p_2p_3=r_1r_2\dots r_{s-1}$ with $r_i$ prime. Repeating this process (assuming $s\geq 3$), we get $1=r_1r_2\dots r_{s-3}$, and all of the $r_i$ are integers, so either $s-3=0$ or $|r_i|=1$. But $r_i\geq 2$, so $s=3$. If $s<3$, then we will instead end up as $p_2p_3=1$ or similar. In this case, $p_i>1$ implies a contradiction. Thus $s=3$ is the only possible conclusion.

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    $\begingroup$ Please see corrected post. $\endgroup$ – Namaste Dec 10 '12 at 2:15
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I might be wrong, but isn't this just an application of the Fundamental Theorem of Arithmetic (FTA)? This theorem states that for any given positive integer, there is only one way to express it as a product of primes. The solution of the problem is then obvious. Of course, for a more complete solution you would have to first prove the FTA, but I won't go into that here..

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If you're worried that the question seems pointless, notice that the corresponding claim for addition is not true, since:

$$p_1 + p_2 + p_3 = q_1 + \cdots + q_s$$

does not require that $s=3$—for example, $$10 + 11 + 12 = 1 + 2 + 3 + 4 + 5 + 18.$$

Even for multiplication the claim is not true unless you require the $p$s and $q$s to be prime:

$$20 \cdot 25 \cdot 30 = 2 \cdot 10\cdot 15 \cdot 50$$

So this claim depends in a crucial way on special properties of multiplication and of prime numbers, and that is why it is not a silly question.

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