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My question is:

Has it been proved/disproved or studied the following?

For every $k\geq 4$ there are $k$ pairwise relatively prime numbers $a_1,a_2,...,a_k$ all greater than $1$ such that $$\text{rad}(a_1+a_2+...+a_k)=\text{rad}(a_1\cdot a_2\cdots a_k)$$

The radical of $n$ is the product of the distinct prime factors of $n$.

Thanks in advance!

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    $\begingroup$ If $k$ is odd, then $1+1+\cdots+1+2+k=1\cdot1\cdots2\cdot k$, where there are $k-2$ ones on each side. $\endgroup$
    – Gerry Myerson
    Commented Nov 10, 2015 at 10:51
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    $\begingroup$ What about 2+2+2+3+3+6=18? $\endgroup$
    – Jan-Christoph Schlage-Puchta
    Commented Nov 10, 2015 at 15:26
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    $\begingroup$ 2048 + 125 + 121 + 9 + 7 = 2310. One might use a density/probabilistic argument to show existence for larger k. Gerhard "But I'm Not Doing That" Paseman, 2015.11.10 $\endgroup$
    – Gerhard Paseman
    Commented Nov 10, 2015 at 18:47
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    $\begingroup$ @Jan-ChristophSchlage-Puchta the numbers you use are not pairwise realtively prime $\endgroup$
    – Konstantinos Gaitanas
    Commented Nov 10, 2015 at 21:44
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    $\begingroup$ I am having a hard time finding a solution with k= 3 a_i for which the radical is 105. This might lead to asking for which squarefree n is a solution with n=rad(...) involving at most k terms possible? Gerhard "Will Let Someone Else Ask" Paseman, 2015.11.12 $\endgroup$
    – Gerhard Paseman
    Commented Nov 12, 2015 at 17:31

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Let $k=2j$ be even, and choose $k$ or more distinct primes, whose product $P$ is the radical of $k$ pairwise coprime numbers to be chosen.

Case $P$ is odd: Then the $2j$ numbers to be chosen are odd, and the sum is even, so $P$ is even, contradiction.

Case $P$ is even: Then, being pairwise coprime, one of the $2j$ numbers is even and the rest are odd, so the sum is odd, so $P$ is odd, contradiction.

So no such solutions for $k$ even.

You can investigate representations of odd squarefree numbers as a sum of powers or special multiples of their factors. I do not know of any examples. I suspect that the subject is a curiosity with few or no ties to other aspects of number theory, yet.

Gerhard "So Go And Make Some" Paseman, 2015.11.10

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    $\begingroup$ Or to paraphrase from dim memory: " But the only number that is both even and odd is infinity, so P is infinite, thus Alexander The Great did not exist and he had an infinite number of limbs." Gerhard "A Random Walk In Science" Paseman, 2015.11.10 $\endgroup$
    – Gerhard Paseman
    Commented Nov 10, 2015 at 18:18
  • $\begingroup$ Note that $P$ can also be the product of $k$ or fewer primes, if one or more instances of 1 are allowed. Thus examples like Jan-Christoph's involving $k$ even must have noncoprime numbers for the $a_i$. Gerhard "Start Looking To Generalize This" Paseman, 2015.11.10 $\endgroup$
    – Gerhard Paseman
    Commented Nov 10, 2015 at 18:59
  • $\begingroup$ What about the example given by Jan-Christoph Schlage-Puchta? $\endgroup$
    – Stefan Kohl ♦
    Commented Nov 10, 2015 at 21:16
  • $\begingroup$ @StefanKohl, perhaps I spoke too briefly. By "noncoprime" I mean at least two of the numbers a_i share a prime factor or more in common. There is a better way of saying this, but it is longer. In any case, the proof above shows that k even means no solution where all integers are coprime in pairs. Jan-Christoph's example has at least two members which share a prime factor. Gerhard "Didn't I Say This Already?" Paseman, 2015.11.10 $\endgroup$
    – Gerhard Paseman
    Commented Nov 11, 2015 at 1:18

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