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Let $n=p_1^{\alpha_1}\cdots p_r^{\alpha_r}$ be the prime decomposition of the integer $n$. Define $$n' = n \sum_{i=1}^r \frac{\alpha_i}{p_i}\quad\text{and}\quad\Omega(n) = \sum_{i=1}^r \alpha_i\quad\text{and}\quad\omega(n) = r.$$ Let $a,b$ be relatively prime, i.e., $\gcd(a,b)=1$, and let $c = a+b$. Suppose that $$\Omega(c) = \min\bigl\{\Omega(a),\Omega(b),\Omega(c)\bigr\}.$$ Is it true that $$\Omega(\gcd(a,a'))+\Omega(\gcd(b,b'))+\Omega(\gcd(c,c')) \le \Omega(ab)-1?$$ From this it would follow that $$\min\bigl\{\Omega(a),\Omega(b),\Omega(c)\bigr\} \le \omega(abc) - 1.$$

Edit: From the answer given by Kevin Buzzard, one can see, that the first inequality is wrong. It is unclear to me however, if the second inequality is also wrong.

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  • $\begingroup$ Is $(a,a')$ the highest common factor? $\endgroup$ Commented Feb 2, 2017 at 19:38
  • $\begingroup$ Yes, the gcd. $(a,a') = gcd(a,a')$ $\endgroup$
    – user6671
    Commented Feb 2, 2017 at 19:39
  • $\begingroup$ Not necessarily. b could divide b' and a could divide a', and no matter how few factors a+b had there would be no chance to satisfy the inequality. Gerhard "Smooth Numbers Can Be Rough" Paseman, 2017.02.02. $\endgroup$ Commented Feb 2, 2017 at 19:43
  • $\begingroup$ Thanks for your comment. Can you make an example? $\endgroup$
    – user6671
    Commented Feb 2, 2017 at 19:49
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    $\begingroup$ Let me guess: you are trying to come up with an elementary proof of the ABC conjecture and you think this "numerical derivative" n' is going to be the key. $\endgroup$
    – KConrad
    Commented Feb 2, 2017 at 22:16

2 Answers 2

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No. How about $a=9$ and $b=16$? Then $c=25$ so $\Omega(a)=\Omega(c)=2\leq\Omega(b)$, the gcd's are $3,16,5$ so the left hand side is 6 and the right hand side only 5.

Edit: if $a=316$ and $b=27$ then we even have the left hand side being greater than $\Omega(ab)$, answering a question in the comments.

Editedit: if $a=544$ and $b=81$ then indeed your "from this it would follow" inequality is wrong.

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    $\begingroup$ Isn't $\Omega(16) = 4$? $\endgroup$
    – user6671
    Commented Feb 2, 2017 at 19:48
  • $\begingroup$ Perhaps this is the first explicit example of the phenomena Gerhard is indicating -- with numbers like 16 you can make $\Omega((b,b'))$ be as big as $\Omega(b)$. $\endgroup$ Commented Feb 2, 2017 at 19:52
  • $\begingroup$ Yes thanks -- slip is now fixed. Answer still OK. $\endgroup$ Commented Feb 2, 2017 at 19:52
  • $\begingroup$ Ok, agreed. Would the inequality hold, if we remove -1? $\endgroup$
    – user6671
    Commented Feb 2, 2017 at 19:54
  • $\begingroup$ Does this mean that the second inequality is also wrong? $\endgroup$
    – user6671
    Commented Feb 2, 2017 at 19:59
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Suppose there are prime numbers $p, q, r$, such that $2^k p+3^kq=5^k r$. Then the left hand side of the second inequality becomes $k+1$, while the right hand side is 5. Thus, in this case, the second inequality does not hold, even if you replace $\omega(abc)-1$ by any function of $\omega$.

To show that such prime numbers actually exist, you use the circle method. The major arcs get a little complicated, as $\sum_{n=1}^N\Lambda(n)e(\alpha 2^k n)$ gets very large at $\alpha=\frac{u}{2^k}$, $u\in\{0, \ldots, 2^k-1\}$, but, thanks to the fact that $2^k$ is not sqaurefree, the other exponential sums become quite small at these points which means that the region close to these points do not contribute significantly.

I would assume that someone already showed that under the obvious conditions on $a, b, c$, the equation $ap+bq=cr$ has the expected number of solutions in primes $p,q,r$, but could not find a reference.

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