Redacted

Here is my solutionl. By : \[(a^2+b^2)(c^2-d^2)=2020.\] We know that $ c^2 - d^2$ is odd or $4 \mid (c^2-d^2)$ and with $2020 = 2^2*5*101$ we get: $(a^2+b^2; c^2-d^2) = (404; 5), (20; 101), (4; 505), (505; 4), (101; 20) (5; 404)$ We left intentionally (2020; 1) because we consider a, b, c, d strictely positive. Note that the number of solution depend on the equation a^2+b^2. <--> $a^2+b^2 = 4 $ $and$ $c^2 - d^2 = 505$ $a^2+b^2 = 4$ no solution because $a; b > 0$ <--> $a^2+b^2 = 20 $ $and$ $c^2 - d^2 = 101$ $a^2 + b^2 = 20 \implies (a,b) = (4; 2) (2; 4)$ $c^2 - d^2 = 101 \implies (c,d) = (51; 50)$ <--> $a^2+b^2 = 404 $ $and$ $c^2 - d^2 = 5$ $a^2 + b^2 = 404 \implies (a,b) = (4; 2) (2; 4)$ $c^2 - d^2 = 5 \implies (c,d) = (3; 2)$ <--> $a^2+b^2 = 5 $ $and$ $c^2 - d^2 = 404$ $a^2 + b^2 = 5 \implies (a,b) = (2; 1) (1; 2)$ $c^2 - d^2 = 404 \implies (c,d) = (102; 100)$ <--> $a^2+b^2 = 101 $ $and$ $c^2 - d^2 = 20$ $a^2 + b^2 = 101 \implies (a,b) = (10; 1) (1; 10)$ $c^2 - d^2 = 20 \implies (c,d) = (6; 4)$ <--> $a^2+b^2 = 505 $ $and$ $c^2 - d^2 = 4$ $c^2 - d^2 = 4 $ no solution because $c;d$ is strictly positive.