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

Question

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.

Answer 1

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 $$ \frac{(x)(2x)(3x)\cdots(nx)}{(ix + jx)}=\frac{x^n n!}{(ix + jx)} = \frac{x^{n-1}n!}{i+j} = \mbox{integer}$$