Folklore

Find all natural $\$ n \$$, satisfying the condition: the numbers $\$ 1,2,3, \backslash d o t s, 2 n \$$ can be paired in such a way that if the numbers in each pair are added and the results are multiplied, the product is a square of a natural number.
$ 1+2 $ - not a square. Let $ n>1 $.

The first method. We will divide these numbers into groups of four consecutive numbers, and, if necessary, a group of six initial numbers. From the groups of four, we form $(a+(a+3))((a+1)+(a+2))=(2a+3)^2$, and from the group of six - $(1+5)(2+4)(3+6)=18^2$.

The second method. If $ n $ is even, then $(1+2n)(2+(2n-1)) \ldots (n+(n+1))=(2n+1)^n$ - a square.

If $ n $ is odd, then $(1+5)(2+4)(3+6)(7+2n)(8+(2n-1)) \ldots ((n+3)+(n+4))=18^2(2n+7)^{n-3}$ - a square.

## Answer

All $ n>1 $.