Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?

Dividing by $(a,b)$, we can assume that $(a,b)=1$.
Multiplying the two equations yields
$$
a^4-b^4=c^2d^2\tag{1}
$$
whereby
$$
b^4+(cd)^2=a^4\tag{2}
$$

This answer characterizes all primitive Pythagorean Triples; that is, there must be a pair of positive integers $(m,n)$ so that $(m,n)=1$ and $m+n$ is odd so that
$$
x=m^2-n^2,\quad y=2mn,\quad z=m^2+n^2\tag{3}
$$
Suppose that $(x,y,z)$ is the Pythagorean triple with the smallest hypotenuse so that at least two sides are a square or twice a square.
Suppose that $z$ and $x$ are squares (since they are odd, they can't be twice a square). Then $(\sqrt{xz},n^2,m^2)$ has a smaller hypotenuse.
Suppose that $z$ is a square and $y$ is a square or twice a square. Then $m$ and $n$ must both be either a square or twice a square. Then $(m,n,\sqrt{z})$ has a smaller hypotenuse (since $x,y\ge1$, $z\gt1$).
Suppose that $x$ is a square and $y$ is a square or twice a square. Then $m$ and $n$ must both be either a square or twice a square. Then $(\sqrt{x},n,m)$ has a smaller hypotenuse.
Thus, there is no primitive Pythagorean Triple with at least two sides a square or twice a square.
If $(2)$ were true, there would be a primitive Pythagorean Triple with two sides a square.
