...for any four real numbers $a,b,c,d$ such that $(a-c)^2-(b-d)^2\geq 0$, we have: $$(a-c)^2-(b-d)^2=(f(a,b)-f(c,d))^2+(g(a,b)-g(c,d))^2$$
The context is that I want to map events on spacetime to transformed co-ordinate axes, such that the space-time interval between any two events becomes equal to the euclidean distance between the two events.
It's okay if multiple events get mapped to the same point. I understand that the points on the line $x=y$ will have to be mapped to the origin.
There are no real functions which have the property you want.
Assume $f,g$ are functions which solve your problem. Then, for $a-c=d-b$, ie $a+b= c+d$, we get : $$0 = f(a,b) - f(c,d) = g(a,b) - g(c,d)$$
this means that there must be functions $F(x),F(y)$ such that : $$f(x,y) = F(x+y)\qquad \text{and}\qquad g(x,y) = G(x+y)$$
But then, for $a=b$ and $c=d=0$, we get : $$0 = F(2a) - F(0) = G(2a) - G(0) $$ so $F,G$ are constant, which is absurd.
Remark : the solution I mentionned originally $f(x,y) = x$ and $g(x,y) = iy$ is used in physics. It is called Wick rotation