1. **Decomposition of Integers:** - Consider each integer \( k \) from \( 1 \) to \( 4n \). - Any integer \( k \) can be expressed in the form \( s \times t^2 \), where \( s \) is square-free (i.e., \( s \) has no square factors other than 1), and \( t \) is a positive integer. - This representation isolates the square-free component \( s \) and the square component \( t^2 \) of each integer. 2. **Pairing Strategy:** - To form pairs whose product is a perfect square, we must pair numbers with the same square-free component \( s \). - For each square-free \( s \), consider the numbers of the form \( s \times (2t - 1)^2 \) and \( s \times (2t)^2 \) for all \( t \). - Pairing these $s \times (2t - 1)^2$ and $s \times (2t)^2$, and thus the number of pairs formed in this way is the largest. 3. **Counting the Number of Pairs:** - For each $s$, as mentioned above, the pairing operation will at most leave one number of the form $s \times (2t + 1)^2$ unpaired. These remaining numbers have the same form as the first term of each pair, $s \times (2t - 1)^2$, and are not multiples of $4$. - Since all multiples of $4$ are already encompassed in a portion of the second element of $s \times (2t)^2$, this constitutes a one-to-one correspondence between all multiples of $4$ and all pairs. - Given that the numbers range from \( 1 \) to \( 4n \), there will be exactly \( n \) such pairs formed. **Conclusion:** For each positive integer \( n \), the maximum number of pairs that can be formed under the given conditions is \( \boxed{n} \).

Some hints ;If you check for some examples, you can easily see; Only $(c,4*c)$ is always solution, that pays all conditions. And c can have these prices $"1,2,..,n"$. ==> $n pairs$

zaidova wrote: Some hints ;If you check for some examples, you can easily see; Only $(c,4*c)$ is always solution, that pays all conditions. And c can have these prices $"1,2,..,n"$. ==> $n pairs$ In this way, the number $4$ both in $(1,4)$ and $(4,16)$.

lbh_qys wrote: zaidova wrote: Some hints ;If you check for some examples, you can easily see; Only $(c,4*c)$ is always solution, that pays all conditions. And c can have these prices $"1,2,..,n"$. ==> $n pairs$ In this way, the number $4$ both in $(1,4)$ and $(4,16)$. i accepted that c can take 1,.., n. Like if n=1 like you said; the numbers are 1,2,3,4 $c={1}$ only. you can look as a subset . for n=4 ${1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}$ $(1,4)$ $(2,8)$ $(3,12)$ $(9,16 )$ i got what do You want to say. another example ${1,2,....,31,32}$ for n=8 (1,4) (2,8) (3,12) (9,16) (5,20) (6,24) (7,28) (18,32) $(c,4c)$ for all $c!=4*k$ and $c={1,2,..,n}$ another solutions are in that form $(9*t,16*t)$ i think now we are done $(c,4*c)$ for $c!=4k$ and $(9*t,16*t)$