Problem 2

Suppose $f(a)=f(b)$. Shifting both $a$ and $b$ by the same large integer $m$ preserves $f(a)=f(b)$ (since $f(a+m)=f(a)-m=f(b)-m=f(b+m)$), and since the discriminant $(a+m+1)^2-4(b+m)$ is a quadratic in $m$ with positive leading coefficient, it becomes nonnegative for $m$ large enough; so we may assume from the start that real numbers $x,y$ exist with $x+y=a+1$, $xy=b$. The original equation then gives $f(f(x)f(y))=f(xy)-f(x+y)=f(b)-f(a+1)=f(b)-(f(a)-1)=1$ (using $f(a)=f(b)$). Applying the shift identity to $u=f(x)f(y)$: $f(u+1)=f(u)-1=1-1=0$, so by the unique zero from Step 3, $u+1=1$, i.e. $f(x)f(y)=0$; by the unique zero again, $x=1$ or $y=1$. If $x=1$ then $y=b$ and $x+y=1+b=a+1$ gives $b=a$; if $y=1$ then symmetrically $a=b$. Either way $a=b$, proving injectivity.