$P(r,b)$ with $r=prime$ givew that $bf(b)$ is a quadratic residu $mod r$ but this true for every $r$ so $bf(b)=g(b)^2$ $P(1,1)$ let $f(1)=y^2$ we have $y^2(y^4+3)=A^2$ esily we get that $f(1)=1$ $P(1,\frac{p-1}{2})$ gives $f(\frac{p-1}{2})=\frac{p-1}{2}$ where $p=odd prime$ Let now $a$ such that $f(a)=a$ and fix $b$ then we have: $(a^2+b)^2+bf(b)-b^2=A^2$ for $a\rightarrow \infty $ we get $bf(b)=b^2\Rightarrow f(b)=b$