Download On integer right triangles with equal area 8

On integer right triangles with equal area
Norbert Hungerbuhler
ETH-Zentrum, Mathematics Department
CH-8092 Zurich, Switzerland
1 Introduction
Right triangles with integer side lengths are one of the earliest traces of human
interest in mathematics. Already in ancient Egypt the right triangle with legs
a = 3 and b = 4, and with hypothenuse c = 5, was known.
As it is well known, and easily checked by the reader, the formulas
(1)
8
>< a = 2mn
2
2
>: bc == ((mm2 ?+ nn2))
give for any choice of nonzero integers m 6= n and an integer right triangle.
(Geometrically we deal with a negative side length as if it were positive.) Such a
triangle is called primitive if gcd(a; b; c) = 1. Obviously, for = 1 and m 6= n not
both odd with gcd(m; n) = 1, the resulting triangle is primitive. On the other
hand, the reader is invited to check, that all primitive triangles are obtained in
this way.
Right triangles with integer side lengths have a lot of interesting properties
and appear in many contexts (see e.g. [1], [2], [7]). Charles Lutwidge Dodgson,
who is better known as Lewis Carroll, the author of \Alice's Adventures in
Wonderland" (1865), was also a many-sided mathematician: In Life and Letters
of Lewis Carroll on page 343 we nd the diary entry
Dec. 19 (Sun).|Sat up last night till 4 a.m., over a tempting problem, sent me from New York, `to nd 3 equal rational{sided rt.{
angled 4's.' I found two , whose sides are 20, 21, 29; 12, 35, 37; but
could not nd three.
Thus Carroll's search was for triples of integer-sided right triangles, all having
the same area. Despite his failure, Carroll conjectured that innitely many
such triples exist. (Two triples are identied if they are equal up to scaling by
a rational factor.) Let us see if we can help Carroll nding such triples.
1
2 Triples of integer right triangles with equal
area
Let us start by trying to nd two integer right triangles in a systematic way.
We make the ansatz:
a1 = 2mn1; b1 = m2 ? n21 ; c1 = m2 + n21
a2 = 2mn2; b2 = m2 ? n22 ; c2 = m2 + n22
for two triangles with sides a1 ; b1 ; c1 and a2 ; b2 ; c2 respectively. Requiring that
both triangles have the same area A means, that
(2)
A = mn1 (m2 ? n21 ) = mn2 (m2 ? n22 )
holds. Dividing (2) by m and collecting terms involving m on the left, we get
(3)
m2 (n1 ? n2 ) = n31 ? n32 (n1 ? n2 )(n21 + n1 n2 + n22 ):
Dividing (3) by n1 ? n2 6= 0 we arrive at
(4)
m2 = n21 + n1 n2 + n22 :
Solving this quadratic equation for n1 we obtain
p
n1 = ? 21 (n2 ? 4m2 ? 3n22 ):
(5)
Hence, the discriminant must be a square number:
(6)
(2m)2 ? 3n22 = 2
for some 2 IIN. The diophantic equation (6) in m; n2 and is a Pell equation
and it is well known (see e.g. [6]), that it has a nontrivial solution (i.e. n2 2=
f0; mg) whenever m is a prime number of the form m = 6N + 1 (or a product
of such numbers). Now we claim, that we have found in fact a nontrivial solution
of the original equation (2).
In fact, since by (6) n2 and are either both even or both odd, (5) implies that
n1 2 ZZ. Furthermore, from (4) we deduce that in fact n1 6= n2 . Hence, we have
a solution of the diophantic equation (2).
Now the following miracle happens. Although we started to search two triangles,
we found in fact three: Remember that, in view of (2), n1 and n2 are two distinct
roots of the third order polynomial p(x) = x3 ? m2 x + mA x(x ? m)(x + m)+ mA .
Hence, there is a third solution n3 which, by Vieta's Theorem (see e.g. [4]),
satises
(7)
n1 + n2 + n3 = 0
(coecient of x2 ) and n1 n2 n3 = ? mA 6= 0. Thus n3 = ?(n1 + n2 ) 2 ZZ and
n1 ; n2 and n3 have distinct absolute values and are dierent from m (consider
2
p(x) = 0). Hence, the integer right triangles Di , i = 1; 2; 3, with sides ai =
2mni , bi = m2 ? n2i and ci = m2 + n2i have common area A = ?mn1n2 n3 and,
since the hypotenuses ci have dierent length, are non-congruent. This triple
has a special property: Notice that (7) implies that at least one of the ni is
even. It is easy to see that non of the ni is a multiple of m. Hence, since m is
chosen to be prime, at least on of the Di is primitive. In particular, the triple
Di , i = 1; 2; 3, is primitive in the sense that the 9 sides do not share a common
factor.
Is it possible that dierent values of m lead to the same triple? The answer is no!
To see this, notice that one of the ni , say n1 , is special by having absolute value
larger then m (look at the graph of p(x)). Thus D1 has maximal hypothenuse
in the triple. Thus, if a1 ; b1 ; c1 are the positive side lengths of D1 , we have that
a1 = 2mjn1j; b1 = n21 ? m2; c1 = n21 + m2 or that a1 = n21 ? m2; b1 = 2mjn1j; c1 =
n21 + m2 . Thus, it follows that
either m =
q c ?a
1
2
1
2 IIN or m =
q c ?b
1
2
1
2 IIN:
Hence, m is determined by the triple.
Thus, for each prime number m = 6N + 1 we nd a primitive triple of integer
right triangles with common area. How many such m exist? Recall that Dirichlet's theorem tells that in every arithmetic progression where consecutive terms
are relatively prime, appear innitely many prime numbers. This theorem assures that there are innitely many prime numbers m = 6N +1. We summarize
our results in the following theorem, which was rst proved in [5]:
Theorem. There exists an innite family of triples of integer-sided right triangles with the same area. In particular, if m is a prime number of the form
m = 6N + 1, N 2 IIN, then there exist positive integer solutions n and l of the
equation m2 = n2 + nl + l2, and a primitive triple of integer-sided right triangles
of area A = mnl(n + l). Dierent values of m lead to dierent primitive triples.
Remark. The method above does not produce all primitive triples. The primitive triple with minimum area which is of a dierent type is
a1 = 4080 b1 = 1001 c1 = 4201
a2 = 1430 b2 = 2856 c2 = 3194
a3 = 528 b3 = 7735 c3 = 7753
generated by (1) using the numbers = 1 and
(m1 ; n1 ) = (51; 40)
(m2 ; n2 ) = (55; 13)
(m3 ; n3 ) = (88; 3) :
The common area is A = 2042040. This example was found by a Mathematica
program.
3
3 Fermat's formula
Apparently, Carroll was not aware of Fermat's observation in [3] that if z is the
hypotenuse and b; d are the legs of a rational right triangle, then we obtain a
new rational right triangle of the same area with sides
4
2 2
4
2 2
2
z 0 = 2zz (+b2 4?b dd2 ) ; b0 = 2zz (?b2 4?b dd2 ) ; d0 = 2z (4bz2 ?bdd2 ) :
Starting with a rational right triangle and iterating this formula n ? 1 times
yields n rational right triangles with equal area and by multiplication with a
suitable integer, the resulting triangles have integer side lengths. However, it
might happen, that some of these triangles are congruent. So, we conclude this
note with the following problem for the reader.
Problem. Prove that for any n 2 IIN there exists an innite family of primitive
n-tuples of integer right triangles with common area.
REFERENCES
1 J. Collins: The Gentleman's Math. Companion, 2, No. 11 (1808), 123.
2 J. Cunlie: New Series of the Math. Repository (ed. Th. Leybourn), 3, II
(1814), 60.
3 P. Fermat: uvres III, 254{255; Fermat's Diophanti Alex. Arith., 1670,
220.
4 M. Hazewinkel: Encyclopaedia of mathematics, Kluwer Academic Publishers, Dordrecht (etc.), 1987.
5 N. Hungerbuhler: Proof of a conjecture of Lewis Carroll, Math. Mag.
69/3, 182-184 (1996)
6 L. J. Mordell: Diophantine equations, Academic Press, London (etc.),
1970.
7 C. Tweedie: Proc. Edinb. Math. Soc., 24 (1905-6), 7{19.
4