No sequential integer squares will ever have a ratio of two. In fact, no integer squares will ever have a ratio of two. That is to say, no integer square will ever be exactly twice the size of another.
Start by assuming $a^2=2b^2$. The quantity $b^2$ has the prime factorization of $list*list$, where "list" is the prime factorization of $b$, multiplied in long form. If $b$ were 10, for example, the element "list" would be replaced with $5*2$.
Because there are two of each prime factor from $b$, we deduce that $b^2$ has an even number of prime factors. The same argument may show that $a^2$ will also have an even number of prime factors.
Obviously, $a^2$ cannot have the prime factorization $primelist*primelist*2$, as this is an odd number of prime factors. Contradiction.