The 35th Japanese Mathematical Olympiad

44 Since $2025=3^4\times 5^2$, each of $a$, $b$, $c$, and $d$ has no prime factors other than $3$ and $5$. Therefore, we can write $$a=3^{x_1}{5}^{y_1},b=3^{x_2}{5}^{y_2},c=3^{x_3}{5}^{y_3},d=3^{x_4}{5}^{y_4}$$ for some non-negative integers $x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4$. Since $abcd=2025$, we obtain $$x_1+x_2+x_3+x_4=4,y_1+y_2+y_3+y_4=2.$$

Also, since $ab$, $bc$, $cd$, and $da$ are all perfect squares, the following sums must be even: $$\begin{gathered} x_1+x_2,x_2+x_3,x_3+x_4,x_4+x_1, \\ {y}_1+y_2,y_2+y_3,y_3+y_4,y_4+y_1. \end{gathered}$$ Thus, all of $x_1,x_2,x_3,x_4$ must have the same parity, and similarly, all of $y_1,y_2,y_3,y_4$ must have the same parity. Conversely, the tuple $(a,b,c,d)$ corresponding to any tuple $(x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4)$ of non-negative integers satisfying (*) and the parity conditions satisfies the given conditions. Therefore, the possible $(x_1,x_2,x_3,x_4)$ are the permutations of $(4,0,0,0)$, $(2,2,0,0)$, $(1,1,1,1)$, and the possible $(y_1,y_2,y_3,y_4)$ are the permutations of $(2,0,0,0)$. Hence, the number of tuples $(a,b,c,d)$ that satisfy the given conditions is $((\left(\genfrac{}{}{0ex}{}{4}{1}\right)+\left(\genfrac{}{}{0ex}{}{4}{2}\right))+1)\times \left(\genfrac{}{}{0ex}{}{4}{1}\right)=44$.