Cryptography Reference
In-Depth Information
1
/
2) and (1
,
1).
Why do we use these points? We are looking for a point of intersection
somewhere in the first quadrant, and the line through these two points seems
to be the best choice. The line is easily seen to be
y
=3
x
Let's repeat the above procedure using the points (1
/
2
,
−
−
2. Intersecting
with the curve yields
(3
x −
2)
2
=
x
(
x
+ 1)(2
x
+1)
6
.
This can be rearranged to obtain
51
x
3
2
x
2
+
···
=0
.
−
(By the above trick, we will not need the lower terms.) We already know the
roots 1
/
2 and 1, so we obtain
1
2
+1+
x
=
51
2
,
or
x
= 24. Since
y
=3
x
−
2, we find that
y
= 70. This means that
1
2
+2
2
+3
2
+
···
+24
2
=70
2
.
If we have 4900 cannonballs, we can arrange them in a pyramid of height 24,
or put them in a 70-by-70 square. If we keep repeating the above procedure,
for example, using the point just found as one of our points, we'll obtain
infinitely many rational solutions to our equation. However, it can be shown
that (24, 70) is the only solution to our problem in positive integers other than
the trivial solution with
x
= 1. This requires more sophisticated techniques
and we omit the details. See [5].
Here is another example of Diophantus's method. Is there a right triangle
with rational sides with area equal to 5? The smallest Pythagorean triple
(3,4,5) yields a triangle with area 6, so we see that we cannot restrict our
attention to integers. Now look at the triangle with sides (8, 15, 17). This
yields a triangle with area 60. If we divide the sides by 2, we end up with
a triangle with sides (4, 15/2, 17/2) and area 15. So it is possible to have
nonintegral sides but integral area.
Let the triangle we are looking for have sides
a, b, c
, as in Figure 1.3. Since
the area is
ab/
2 = 5, we are looking for rational numbers
a, b, c
such that
a
2
+
b
2
=
c
2
,
ab
=10
.
A little manipulation yields
a
+
b
2
2
=
c
2
2
+5
,
=
a
2
+2
ab
+
b
2
4
=
c
2
+20
4
a
−
b
2
2
=
c
2
2
=
a
2
−
2
ab
+
b
2
4
=
c
2
−
20
4
−
5
.
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