$f\left(x\right)=x^2-2ax+a\left(a+1\right),f:\left[a,\infty \right)\to \left[a,\infty \right).$ If one of the solutions of the equation $f\left(x\right)=f^{-1}\left(x\right)$ is 5049, then the other may be

$\begin{array}{l}f\left(x\right)=x^2-2ax+a\left(a+1\right)\\ f\left(x\right)={\left(x-a\right)}^2+a,x\in \left[a,\infty \right)\\ \mathrm{Let} y={\left(x-a\right)}^2+a. \mathrm{Clearly}, y\ge a. \mathrm{Thus},\\ {\left(x-a\right)}^2=y-a\\ \mathrm{or} x=a+\sqrt{y-a}\\ \therefore f^{-1}\left(x\right)=a+\sqrt{x-a}\\ \mathrm{Now}, f\left(x\right)=f^{-1}\left(x\right)\\ \mathrm{or} {\left(x-a\right)}^2+a=a+\sqrt{x-a}\\ {\left(x-a\right)}^2=\sqrt{x-a}\\ \mathrm{or} {\left(x-a\right)}^4=\left(x-a\right)\\ \mathrm{i}.\mathrm{e}.,x=a \mathrm{or} {\left(x-a\right)}^3=1\\ \mathrm{i}.\mathrm{e}., x=a \mathrm{or} a+1\\ \mathrm{If} a=5049, \mathrm{then} a+1=5050\\ \mathrm{If} a+1=5049, \mathrm{then} a=5048\end{array}$