Almost Lagrange's Four Squares

Let ${a}^2 + {b}^2 = \alpha$

$\implies {c}^2 + {d}^2 = X - \alpha$

Now, we will create a hash-table $d$ such that,

$d$ = {A particular Value of $\alpha$ : Number of ways it can be represented as sum of two perfect squares}

So for example, $d[\alpha = 50]$ = $2$ , as 50 can be represented in two ways as sum of two squares, $50$ = ${1}^2 + {7}^2$ = ${5}^2 + {5}^2$

Now, all we have to do is generate all possible values of $\alpha$ and store it in $d$ .

Now, we iterate through each and every element in $d$ , and check if $X - \alpha$ is there in the table(checking existence of an element in a hash table takes O(1) time)

If $X - \alpha$ doesn't exist, then $X$ cannot be represented as sum of four squares for that particular value of $\alpha$ . If $X - \alpha$ does exist, then number of ways to write $X$ as sum of four squares becomes,

$d[\alpha] * d[X - \alpha]$

Using this logic, and little bit of optimization, your code can run in less than 20s