Cryptography Reference
In-Depth Information
The tangent line is therefore
y
=
23
12
x
+
41
3
.
Intersecting with the curve yields
23
2
12
x
+
41
=
x
3
−
25
x,
3
which implies
23
12
2
x
3
x
2
+
−
···
=0
.
Since the line is tangent to the curve at (
−
4
,
6), the root
x
=
−
4 is a double
root. Therefore the sum of the roots is
4+
x
=
23
12
2
−
4
−
.
We obtain
x
= 1681
/
144 = (41
/
12)
2
. The equation of the line yields
y
=
62279
/
1728.
Since
x
=(
c/
2)
2
,weobtain
c
=41
/
6. Therefore,
=
y
=
(
a
2
b
2
)
c
=
41(
a
2
b
2
)
62279
1728
−
−
.
8
48
This yields
− b
2
=
1519
36
a
2
.
Since
a
2
+
b
2
=
c
2
=(41
/
6)
2
,
we solve to obtain
a
2
= 400
/
9and
b
2
=9
/
4.
We obtain a triangle (see
Figure 1.4) with
a
=
20
b
=
3
c
=
41
6
,
which has area 5. This is, of course, the (40
,
9
,
41) triangle rescaled by a factor
of 6.
There are infinitely many other solutions. These can be obtained by suc-
cessively repeating the above procedure, for example, starting with the point
just found (see Exercise 1.4).
The question of which integers
n
can occur as areas of right triangles with
rational sides is known as the
congruent number problem
. Another for-
mulation, as we saw above, is whether there are three rational squares in
arithmetic progression with difference
n
. It appears in Arab manuscripts
around 900 A.D. A conjectural answer to the problem was proved by Tunnell
in the 1980s [122]. Recall that an integer
n
is called squarefree if
n
is not
3
,
2
,
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