If $f(a;f(b;c)) = f(a;b) + c$ and $f(a;a) = 0$ , then find $f($1.1$; -5)$

Question

Function $f(x;y)$defined on pairs of real numbers satisfies the conditions $f(a;a) = 0$, $f(a;f(b;c)) = f(a;b) + c$ for any $a, b, c$... Find$f($1.1$; -5)$

Let me think first , $f(a;a) = 0 $ for any real numbers. Now it can be said that:

=> $f(1.1;-5)+c = f(1.1;f(-5;c))$

=> $f(1.1;-5)= f(1.1;f(-5;c))-c$

I lost my path waht to do next!! Please help me with the explanation and solution.

Answer 1

Hint: The path to victory is always to try and manipulate variables so that you can get something you know. Since we know something about $f(b;b)$, consider $f(a; 0) = f(a; f(b; b)) = f(a; b) + b$, which has to be constant as $b$ varies. This ultimately lets you find a closed form for $f(0; c)$.

Once you've got that, consider $f(0; f(x; y))$.