The number of points with integral coordinates that lie in the interior of the region common to the circle $${x^2} + {y^2} = 16$$   and the parabola $${y^2} = 4x$$  is :

Solution :
$$\left( {\lambda ,\,\mu } \right)$$  is interior to both the curves if $${\lambda ^2} + {\mu ^2} - 16 < 0$$    and $${\mu ^2} - 4\lambda < 0$$
Now, $${\mu ^2} - 4\lambda < 0\,\,\,\, \Rightarrow \lambda > {\left( {\frac{\mu }{2}} \right)^2}$$
Hence, if $$\mu = 0,\,\lambda = 1,\,2,\,3,\,.....;$$     if $$\mu = 1,\,\lambda = 1,\,2,\,3,\,.....;$$     if $$\mu = 2,\,\lambda = 2,\,3,\,.....;$$     if $$\mu = 3,\,\lambda = 3,\,4\,.....;$$
Also $${\lambda ^2} + {m^2} - 16 < 0\,\,\,\,\, \Rightarrow {\lambda ^2} < 16 - {\mu ^2}$$
Hence, if $$\mu = 0,\,\lambda = 1,\,2,\,3\,;$$    if $$\mu = 1,\,\lambda = 1,\,2,\,3\,;$$    if $$\mu = 2,\,\lambda = 2,\,3\,;$$    if $$\mu = 3,\,\lambda $$  has no integral value.
$$\therefore \,\,\left( {1,\,0} \right),\,\left( {2,\,0} \right),\,\left( {3,\,0} \right),\,\left( {1,\,1} \right),\,\left( {2,\,1} \right),\,\left( {3,\,1} \right),\,\left( {2,\,2} \right),\,\left( {3,\,2} \right)$$           are the possible points.