Changing factorization (2)

If the roots of a quadratic equation are both integers, the discriminant is a perfect square. So let $a^2+4b=m^2$ and $a^2-4b=n^2$ . Then, $(a^2+4b)+(a^2-4b)=m^2+n^2\\m^2+n^2=2a^2$ As $m$ and $n$ have same parity, let $m=u+v$ and $n=|u-v|$ . Then, $(u+v)^2+(|u-v|)^2=2a^2\\u^2+2uv+v^2+u^2-2uv+v^2=2a^2\\2u^2+2v^2=2a^2\\u^2+v^2=a^2$ So $u,v,a$ is a Pythagorean triple. Let $u=k(r^2-s^2),v=2krs,a=k(r^2+s^2)$ ( $k,r,s$ are positive integers, $r>s$ , $\gcd(r,s)=1$ and $2\nmid r+s$ ).

As $m^2-n^2=(a^2+4b)-(a^2-4b)=8b$ , $b=\dfrac{m^2-n^2}{8}=\dfrac{(u+v)^2-(u-v)^2}{8}=\dfrac{(u^2+2uv+v^2)-(u^2-2uv+v^2)}{8}=\dfrac{4uv}{8}=\dfrac{uv}{2}=\dfrac{2k^2rs(r^2-s^2)}{2}=k^2rs(r^2-s^2)$ As $\gcd(a,b)=\gcd(k(r^2+s^2),k^2rs(r^2-s^2))=1$ , $k=1$ .

After trying (trying from $r+s=3,5,7,\cdots$ ), we find that when $(r,s)=(5,2)$ or $(6,1)$ , $b=210$ , which is the minimum requirement.