# Find four distinct positive integers whose product is divisible by the sum of every pair of them.

The teacher gave us British Mathematical olympiad $$1992$$ Round $$1$$ Problem $$3$$.

Find four distinct positive integers whose product is divisible by the sum of every pair of them. Can you find a set of five or more numbers with the same property?

I can't do the question, and my friends don't know either.

Can someone help me? Any help is appreciated.

• Try: 4, 12, 20, 28. Aug 10, 2019 at 2:54
• Also: 6, 14, 22, 26, 30 Aug 10, 2019 at 3:08
• Find an $N$ for which $N,2N,3N,4N$ works. Aug 10, 2019 at 3:19
• $(2a,6a,10a,14a)$ works for all $a$ Aug 10, 2019 at 3:22
• $13!,2\times13!,3\times13!,\dots,8\times13!$. Aug 10, 2019 at 4:16

Summarizing the results of the comments (where $$a$$ is an arbitrary positive integer):
• 4 numbers: $$(2a,6a,10a,14a)$$ [from Nilotpal]
• 5 numbers: $$(6a,14a,22a,26a,30a)$$ and $$(2a,6a,10a,14a,18a)$$.
• 6 numbers: $$(6a,14a,22a,26a,30a,34a)$$ and $$(2a,6a,10a,14a,18a,22a)$$
Generalizing the Empy2 method to show it can work for any number of integers $$n \geq 2$$:
• $$n$$ numbers: $$(x, 2x, 3x, ...., nx)$$ where $$x = \prod_{(i,j) \in D_n} (i+j)$$, where $$D_n$$ is the set of all distinct pairs $$(i,j)$$ for $$i,j \in \{1, ..., n\}$$ and $$i< j$$. Indeed, choosing any $$ix$$ and $$jx$$ for $$(i,j) \in D_n$$ we get $$\frac{(x)(2x)(3x)\cdots(nx)}{(ix + jx)}=\frac{x^n n!}{(ix + jx)} = \frac{x^{n-1}n!}{i+j} = \mbox{integer}$$
• The choice of $x$ in the general case above gives in an easy proof, but it is often larger than necessary. Of course, if $(y[1],...,y[n])$ works then $(ay[1], ay[2], ..., ay[n])$ also works (for any positive integer $a$) as emphasized in the Nilotpal comment. Aug 10, 2019 at 5:03