$$ 2b^2+ab=(a+b)^2+b$$$$ b(2b+a-1)=(a+b)^2$$Suppose that exist $0\equiv b\equiv 2b+a-1\equiv a-1\pmod p$ then $b\equiv a-1\pmod p$ but $b\mid a^2$ then $p\mid a^2$ then $1\equiv 0\pmod p$, contradiction. Therefore $(b,2b+a-1)=1$ and $b$ is a square