Call the numbers $a,b,c,d,e$. The gcd of two distinct numbers can be $1,2,3,4$. Hence if we delete all factors 2,3, then the resulting numbers $p,q,r,s,t$ are pairwise coprime. Moreover the lcm of these numbers is also a square, since a prime $\pi>5$ can appear in at most one of the numbers $a,b,c,d,e$ and then with an even exponent. It follows that $p,q,r,s,t$ are all squares. We have $a\ge 2$ hence $\{a,b,c,d,e\}$ can contain at most one square. Hence at most one of these numbers can be coprime to 6. That means $a\equiv 0,2(6)$. - $a\equiv 0(6)$. Then $b\equiv 1(6)\implies (b,6)=1\implies b=\Box \implies b\equiv 1(4)\implies d\equiv 3(4)\implies d=3^k\Box$. Then $k$ is odd $\implies \nu_3(a)\equiv 0(2)$. Since $b$ is a square, $a$ is not and $\nu_2(a)\equiv 1(2)\implies \nu_2(e)\equiv 0(2)$, because $a,e$ are 0 mod 4 and the maximal number of factors 2 appears in $a$ or $e$. But $e=2^*\Box $ hence $e$ is a square, contradiction. - $a\equiv 2(6)$. Then $d\equiv 5(6)\implies (d,6)=1\implies d=\Box \implies d\equiv 1(4)\implies c\equiv 0(4)$ and $\nu_2(c)$ is even. We have $2|c, 3\not|c$ hence $c=2^*\Box\implies c=\Box$, contradiction.