19.1 .37 Give methods to find all pairs of positive integers, the sum and product of each pair being square numbers. In particular, find all such pairs of positive integers less than 100.
Let $a, b$ be any positive integers, $c^{2}$ denote the largest square factor of $a^{2}+b^{2}$, then $a^{2}+b^{2} = k c^{2}$. Let $x=a^{2} k, y=b^{2} k$, then $x+y=a^{2} k+b^{2} k=(a^{2}+b^{2}) k=(k c)^{2}$, and $x y=(a b k)^{2}$.

We prove: All pairs of positive integers whose sum and product are square numbers can be obtained by appropriately selecting $a$ and $b$ using the method described above.

Assume $x+y=z^{2}, x y=t^{2}$, where $z, t$ are positive integers, let $d=(x, y), c_{1}$ be the largest square factor of $d$, then $d=k c_{1}^{2}$, where $k$ is a square-free positive integer. Thus, $x=d x_{1}, y=d y_{1}$, where $(x_{1}, y_{1})=1$. From $x+y=z^{2}$ we get $(x_{1}+y_{1}) d=z^{2}$, thus $d=k c_{1}^{2} \mid z^{2}$, and since $k$ is square-free, $k c_{1} \mid z$, so let $z=k c_{1} z_{1}$, where $z_{1}$ is a positive integer, thus $(x_{1}+y_{1}) d - x+y = z^{2} = k^{2} c_{1}^{2} z_{1}^{2} = k d z_{1}^{2}$, hence $x_{1}+y_{1}=k z_{1}^{2}, x_{1} y_{1}=t^{2} / d^{2}$, and since $(x_{1}, y_{1})=1$, $x_{1}, y_{1}$ are both square numbers, i.e., $x_{1}=a_{1}^{2}, y_{1}=b_{1}^{2}$, thus $x=d x_{1}=k(c_{1} a_{1})^{2}, y=d y_{1}=k(c_{1} b_{1})^{2}$. Let $a=c_{1} a_{1}, b=c_{1} b_{1}$, we get $x=k a^{2}, y=k b^{2}$, and $a^{2}+b^{2}=(c_{1} a_{1})^{2}+(c_{1} b_{1})^{2}=c_{1}^{2}(x_{1}+y_{1})=k(c_{1} z_{1})^{2}$. Let $c_{1} z_{1}=c$, we get $a^{2}+b^{2}=k c^{2}$. Since $k$ cannot be divisible by the square of an integer greater than 1, $c^{2}$ is the largest square factor of $a^{2}+b^{2}$.

Pairs of positive integers not exceeding 100 whose sum and product are square numbers are: 2,2; 5,20; 8,8; 10,90; 18,18; 20,80; 9,16; 32,32; 50,50; 72,72; 2,98; 98,98; 36,64.