Complete solutions were sent in by Ruth from Manchester High School for Girls, by Ben who did not name his school and by Andrei from Tudor Vianu National College, Romania.

This is Ben's solution to the first part:

Suppose (x1, y1) is a point on the parabola y = ax2 where x1 and y1 are integers, that is y1=ax12. Then, if n is an integer, nx1 and n2 are also integers and so
a(nx1)2 = n2(ax12) = n2y1.
So (nx1, n2y1) is another solution with integer coordinates. As n can take an infinite number of integer values, if there is at least one lattice point solution, there are an infinite number.

This is Ruth's solution to the second part:

On the hyperbola x2 -y2 = 84 = (x+y)(x-y), for x and y to be integers, (x+y) and (x-y) have to be the same parity because (x+y)+(x-y)=2x and, for the total of two numbers to be even, they either have to be both odd or both even.

As (x+y)(x-y)=84 at least one bracket has to be even. As we require diophantine solutions, both brackets must be even. The only factorisations of 84 into two even numbers are:
84=2×42 = 42 ×2 = -2×-42 = -42 ×-2 = 6×14 = 14×6 = -6×-14 = -14×-6.
Each of these gives a distinct solution so the 8 solutions are x = ±22, y=±20 (4 solutions) and x=±10, y=±4 ( 4 solutions).

There are two lattice points on the hyperbola in the first quadrant: (10,4) and (22,20). The lattice points (10, -4) and (22, -20) are the reflections of these points in the x-axis. Also x ³ Ö84 or x £ -Ö84 so there are two branches of the hyperbola. The other four lattice points lie on the other branch of the hyperbola and are the reflections of these four points in the y-axis.