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'Lattice Points' printed from http://nrich.maths.org/
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,
This is Ben's solution to the first
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
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.