Copyright © University of Cambridge. All rights reserved.

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 $(x_1, y_1)$ is a point on the parabola $y = ax^2$ where $x_1$ and $y_1$ are integers, that is $y_1=ax_1^2$. Then, if $n$ is an integer, $nx_1$ and $n^2$ are also integers and so $$a(nx_1)^2 = n^2(ax_1^2)= n^2y_1.$$ So $(nx_1, n^2y_1)$ 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 $x^2 -y^2 = 84 = (x+y)(x-y)$, 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: $$84=2\times 42 = 42 \times 2 = -2\times -42 = -42 \times -2 = 6\times 14 = 14\times 6 = -6\times -14 = -14\times -6.$$ Each of these gives a distinct solution so the 8 solutions are $x= \pm 22, y=\pm 20$ (4 solutions) and $x=\pm 10, y=\pm 4$ ( 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 $x\geq \sqrt 84$ or $x\leq -\sqrt 84$ 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.

This is Ben's solution to the first part:

Suppose $(x_1, y_1)$ is a point on the parabola $y = ax^2$ where $x_1$ and $y_1$ are integers, that is $y_1=ax_1^2$. Then, if $n$ is an integer, $nx_1$ and $n^2$ are also integers and so $$a(nx_1)^2 = n^2(ax_1^2)= n^2y_1.$$ So $(nx_1, n^2y_1)$ 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 $x^2 -y^2 = 84 = (x+y)(x-y)$, 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: $$84=2\times 42 = 42 \times 2 = -2\times -42 = -42 \times -2 = 6\times 14 = 14\times 6 = -6\times -14 = -14\times -6.$$ Each of these gives a distinct solution so the 8 solutions are $x= \pm 22, y=\pm 20$ (4 solutions) and $x=\pm 10, y=\pm 4$ ( 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 $x\geq \sqrt 84$ or $x\leq -\sqrt 84$ 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.