Suppose that this is possible. All of the numbers have to be of the form $k^2,2k^2,3k^2,6k^2$ as $2$ and $3$ are the only prime numbers that can have odd exponents in the prime factorization of each of the $5$ numbers. (this is because the only prime common divisor is 2 and 3) By the pigeon hole principle, there must be at least two numbers with the same form. so we have $a(x^2-y^2)<4$ This only holds when $x=2,y=1$ and $a=1$ and checking $lcm (1,2,3,4,5)=60$, which is not a square number. Contradiction Therefore it is not possible