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It's been a long time since I've posted on here, but a friend of mine recently observed something in number theory and wants to know if anyone can help prove or disprove it, the conjecture is as follows. He's foreign, I apologize for the awkward formatting. I would format it myself, but I'm in a bit of a rush and cannot re-learn how to do so.

For reference, he has defined his "$\text{rad}(x)$" function as "the minimum of the unique prime factors of $x$." $a$ and $b$ are assumed to be positive integers.

$$\begin{array}{c}\large \mathbf{ \text{ Papava's conjecture}} \\ \\ \text{If }a+b=c>2 \text{ and }\gcd(a,b)=1\\ \text{then}\\ c<\gcd(abc, (\text{rad}(abc))^3)\\ \\ \end{array}$$

EDIT:

The OP's definition of $\text{rad}(x)$, as given above, is surely not the one used by the actual proposer of the conjecture (else there are lots of instant counterexamples). The definition which is both standard and also makes sense is this:

For a positive integer $x$, define $\text{rad}(x)$ to be the product of the distinct prime factors of $x$.

With that definition, the conjecture at least survives testing with small numbers.

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  • $\begingroup$ what has been tried ? $\endgroup$
    – user451844
    Aug 31, 2017 at 18:08
  • $\begingroup$ As the text in the graphic is not formatted either, the image is not even an advantage over just typing without formatting, is it? $\endgroup$ Aug 31, 2017 at 18:08
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    $\begingroup$ I'm not sure what he's tried. He's a guy on the Facebook page "math" who has been spamming us with this question and others for a while now. He seems to believe he's some kind of super genius, referring to himself as the "king of number theory." And Hagen, perhaps not. $\endgroup$ Aug 31, 2017 at 18:10
  • $\begingroup$ Is this a consequence of the ABC conjecture? $\endgroup$ Aug 31, 2017 at 18:24
  • $\begingroup$ @GTonyJacobs Somewhat. The ABC-conjecture with $\epsilon=2$ says $c\le \operatorname{rad}(abc)^3$ for almost all triples. However, the gcd may cut this down, in particular if $a,b$ are square-free and $c$ has some high-power factors $\endgroup$ Aug 31, 2017 at 19:54

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Assuming that $\text{rad}(16x)=2$, it looks like $a=5, b=11$ is a counterexample:

$c=16$
$\text{rad}(abc)=2$
$\text{rad}(abc)^3=8$
$\gcd(abc, 8) = 8 < 16$


If instead $\text{rad}(x)$ indicates the largest square-free factor of $x$ - the product of all its prime factors - giving eg. $\text{rad}(144)=6$ we could investigate the sum and difference of prime cubes (or higher powers), looking for something with lower prime factors.

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  • $\begingroup$ It's $rad(abc) = rad(5*11*16) = 2*5*11$ $\endgroup$
    – Paul LeVan
    Aug 31, 2017 at 18:51
  • $\begingroup$ @PaulLeVan OK, I didn't get that from the definition. That would be more interesting, thanks for the alternative interpretation. $\endgroup$
    – Joffan
    Aug 31, 2017 at 18:58
  • $\begingroup$ @PaulLeVan the question indicates that his "rad" means the smallest prime factor of a number $\endgroup$
    – Will Jagy
    Aug 31, 2017 at 19:02
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    $\begingroup$ I was going off what the radical of an integer usually means. If it isn't this normal definition, then yes it is simple to disprove. I would suspect the wording is incorrect. $\endgroup$
    – Paul LeVan
    Aug 31, 2017 at 19:15
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    $\begingroup$ @Joffan: I edited the question to reflect the standard meaning of $\text{rad}(x)$. With the OP's stated definition, the problem is silly. With the corrected definition, counterexamples won't be so easy to find. $\endgroup$
    – quasi
    Aug 31, 2017 at 19:48

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