Multiple Pythagorean Number Triples

Multiple Pythagorean Number Triples

Author(s): Albert Fassler

Source: The American Mathematical Monthly , Jun. - Jul., 1991, Vol. 98, No. 6 (Jun. - Jul., 1991), pp. 505-517

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Multiple Pythagorean Number Triples

ALBERT FASSLER*

ALBERT FASSLER obtained his bachelor's degree in Mechanical Engineering

at the HTL in Winterthur before he studied Mathematics at the Eidgenossische Technische Hochschule ETH in Zurich (Federal Institute of

Technology). After graduation, he worked under Eduard Stiefel at the Seminar ffur Angewandte Mathematik (Seminar for Applied Mathematics) at the ETH, where he got his Ph.D. and became a lecturer. Together with

Stiefel, he wrote a book in the field of applied group theory with the title Gruppentheoretische Methoden und ihre Anwendung (Teubner, Stuttgart, 1979). He is now at the HTL Biel/Bienne (an Engineering college).

SUMMARY OF THIS ARTICLE. We focus on the following questions: How many Pythagorean number

triples are there, whose triangles have a common hypotenuse, leg-sum, leg-difference, area, perimeter,

inradius or leg, and how can they be calculated? How are such triples distributed for small parameters

as well as asymptotically? Statistical results generated by computer will be given. Some of the asymptotic statements may appear surprising to readers who are relatively unfamiliar with number

theory. A survey of published results and historical remarks are included in this article to make it more

enjoyable.

1. Introduction. A triple of strictly positive integers x, y, h satisfying

X2 + Y2 = h 2(1.1)

is called a Pythagorean number triple, or more briefly a Pythagorean triple. A Pythagorean triple, denoted by [x, y, hi, is called primitive if and only if

x, y, h are relatively prime, i.e. their gcd is 1; it is sufficient that gcd(x, y) = 1 holds.

If x and y were both odd then (1.1) would imply that h2 2 (mod 4). But this is impossible, because if h is even this implies h2 0 O (mod 4) and if h is odd this implies h2 = 1 (mod 4). Consequently, one leg has to be even and the other has to be odd. The following way of describing primitive Pythagorean triples (ppt) has been known at least since the time of Euclid.

The pairs of integers [a, b] with the properties

a>b>O gcd(a,b)=1 a+b-1 (mod2) (1.2)

generate the set of all ppt [x, y, h] with x even and y odd by taking

x = 2ab y = a2 - b2 h = a2 + b2. (1.3)

The above description is a bijection between the pairs [a, b] with properties (1.2) and all the ppt [x, y, h] with even x.

If we use any pair of integers a, b with a > b > 0 in (1.3), we would get a Pythagorean triple, but it would not be primitive in general. We call a and b the

generators of the primitive triple [x, y, h].

*Institute for Mathematics and its Applications, University of Minnesota, Minneapolis, and Ingenieurschule Biel, Postfach 1180, CH-2500 Biel/Bienne Switzerland. This work was mainly done when the author was spending his sabbatical in Minneapolis. He wishes to thank the State of Bern, in Switzerland, for its support and the IMA, at the University of Minnesota, for its provision of excellent working conditions and computer facilities.

505



506 ALBERT FASSLER [June-July

Examples:

[2, 1] - [4,3,5] [3,2] - [12,5, 13] [21,4] - [168,425,457].

GEOMETRIZATION. (1.3) can be considered as the conformal mapping z - z2 with z = a + ib:

(a + ib)2 = (a2 - b2) + i * 2ab. (1.4)

Indeed, y = Re(z 2) and x = Im(z2). Introducing coordinate axes as shown in

FIGURE 1, the mapping z - z2 has the following properties when discussed for the entire (a, b)-plane:

y=a2 _b2 80- /(36,77,85)

70--- 8 (16,63,65) (12,65,97)

60-

(48,55,73) '

50-II/\

X - 1~~~~(28,45.53) /

40-- / 14 (80,39,89) a G_ 12,35,37) // \

/56, 33,65)

ot(4 , X 5 )/t < < X XX~~~x 2ab

0 10 20 30 40 50 60 70 80

FIG. 1

For a = a0 = const, we get y = aO - b2 and x = 2a0b. For a0 # 0 this implies b = x/2a20 so that

Y 4a2 + aO (1.5)

As a result, the images of horizontal lines are parabolas which are symmetric about

the y-axis, and which open downward, with their common focus at the origin 0.

The exceptional case of a0 = 0 gives us the negative y-axis as the image.



1991] MULTIPLE PYTHAGOREAN NUMBER TRIPLES 507

Similarly, the images of vertical lines b = = const are described by

Y 4b -ba, bo 0 ? (1.6)

and they are parabolas that open upwards.

Furthermore, y = 0 implies x2 = 4a4 or x = 4b4, respectively. Consequently, parabolas of the two orthogonal families have a point on the positive x-axis in

common if and only if ao = bo. FIGURE 2 shows the images (x, y) of all points (a, b.) with a > b > 0, a -

1, 2, 3,... ,42 and b = 1, 2, 3,.. ., 30. In particular, the computer generated points

mark all ppt in the range x and y < 1800, where the distance from the origin to such a point gives the length of the hypotenuse of the corresponding right triangle. The points marked with an h correspond to primitive multiple triples with a

common hypotenuse. Similarly, s stands for a common leg-sum and f for a common area. See the following for details.

1750 42

1500 .

1250

1000 2

750

500D

250 A

02b -0 250 500 750 1000, 1250 1500 1750 2000

FIG. 2

Note that

_ (p(a) if ais even Number of points on a curve with const a-value = (a)2 If a is odd, (1.7)

(p(a)12 if a iS odd,

where (p(a) = number of positive integers less than a and relatively prime to a. That is, f is the Euler-phi function. If a is even, (1.7) holds by definition. If a is odd, we use the fact that gcd(a, k) = gcd(a, a - k) holds for every k. But k and a - k have different parity. Therefore, exactly one of a pair k, a - k is even.




If by reflection with respect to the line x = y in FIGURE 2, we map all the points with y > x into the domain x > y > 0, we get FIGURE 3. The density of ppt

doubles if we drop the condition that x must be even, however we lose the nice geometrical pattern.

Please note: We always identify triangles that are mirror images of each other.

0 250 1750

FIG. 3

2. Triples with Common Hypotenuse h. If gcd(a, b) = 1, then any odd prime divisor of h = a2 + b2 is of the form 4m + 1. See [4, Theorem 13].

Because h is odd, it is necessary that its prime factorization have the form

h= 1 .22 ..pn (2.1) with pairwise different primes pi - 1 (mod 4) and ,3i _ 1.

Fermat established that each prime number p of the form 4m + 1 can be written uniquely as a sum of two squares: p = a2 + b2, up to a permutation of the squares. We know from the above discussion that a, b have different parity.

When discussing representations of a positive integer as a sum of two squares, we will need the following two identities:

/ (r2 + S2) * (U2 + V2) = (ru - SV)2 + (rv + SU)2 (2.2) (r2 + s2) * (U2 + V2) = (ru + SV)2 + (rv - su)2.

Consequently, if we can represent each factor as a sum of two squares, (2.2) gives us their product as a sum of two squares.

Furthermore, let us call the representation of a number as a sum of two squares

a2 + b2 proper, if gcd(a, b) = 1. With the knowledge of the representation for the primes pi in (2.1), the unique proper representations for the prime powers pfz as well as the total set of 2n- 1 different proper representations for h can be generated by the composition (2.2): Each additional prime power pfz doubles the number of proper representations. Fermat (1632-1690), Euler (1707-1783), Lagrange (1736-1813) and Legendre (1752-1833) played important roles in the development of these ideas (see [18, Chap. II, ?VIII and ?IX] and [3, Chap. VI]). The parity condition is respected by the two compositions (2.2).




In summary, we have the following:

THEOREM 1. If h > 1 has the form (2.1), then there are exactly 2'-1 different primitive Pythagorean triples with common hypotenuse h, which can be generated by applying (2.2). If h is not of the form (2.1), then there is no primitive Pythagorean triple.

Examples: The smallest h with exactly (a) 1 ppt is 5 = 22 + 12 and the triple is [4,3,5] (b) 2 ppt is 5 - 13 = 65 = (22 + 12)(32 + 22) = (6 ? 2)2 + (4 } 3)2 = 82 +

12 = 42 + 72 and the triples are [16,63,65] and [56,33,65]. (c) 4 ppt is 65 * 17 = 1105. From [8, 1], [7, 4] for 65, and using [4, 1] for 17, we

get:

[33,4] -*[264,1073,1105] [31,12] - [744,817,1105]

[32,9] * [576,943, 1105] [24,23] +->[1104,47, 1105]

(d) 128 ppt is 5 * 13 - 17 - 29 * 37 * 41 - 53 * 61 = 138,863,452,745. The initial elements of the ordered sequence of h-values with (1) 4 ppt are: 1105 1885 2405 2465 2665 3145 3445 3485 3965 4505 4745 5185.... Only the first two, marked with h, are shown in FIGURE 2.

(2) 8 ppt are: 32,045 40,885 45,305 58,565 67,405 69,745 77,285 80,665....

DISTRIBUTION OF TRIPLES WITH COMMON HYPOTENUSE.

The asymptotic behavior of the number r(x, m, r) of all primes p < x with p r (mod m) and gcd(r, m) = 1, as x -> oo, the so called primes in arithmetic progression, (see [10, ?4, Ch. III, Th. 3.17]) is given by:

1 x 7r(x, m, r) . . (2.3)

p0(m) log x

In our case, with the notation wr(x) for the number of primes < x, we get

1 x 1 w(x,4,1) 2- _ - I 27r(x). (2.4)

2 log x 2

To generate h-values with several ppt, one has to deal with primes of the form 4m + 1. TABLE 1 shows some values of r(x) and ir(x, 4, 1).

TABLE 1. Distribution of primes.

x T(x) -(x,4,1)

100 25 11

1,000 168 80

2,000 303 147

4,000 550 269

8,000 1007 499

16,000 1862 920

32,000 3432 1705




We let Hj(x) be the number of h ? x such that there exist exactly 2'-1 different ppt [x, y, h] for each h. Computer calculations resulted in TABLE 2 and FIGURE 4.

TABLE 2. Distribution of multiple ppt with common hypothenuse.

x H2(x): 2 triples H3(x): 4 triples H3(x): 8 triples

10,000 100 10 0

20,000 183 26 0

40,000 338 35 1

100,000 664 102 4

200,000 1,183 170 5

500,000 2,429 363 20 1,000,000 4,184 654 58 2,000,000 7,608 1,251 78 4,000,000 13,236 1,943 174 10,000,000 28,624 4,090 522 20,000,000 51,574 7,284 707 50,000,000 111,773 15,203 1,718 100,000,000 204,891 28,300 3,480 200,000,000 369,064 44,444 6,401 400,000,000 680,850 79,373 11,138 1,000,000,000 1,519,597 161,274 22,640 2,000,000,000 2,803,231 301,286 35,707

106. H2(x) o31 C~~~~~~~~~~(X) 12

106

1 _ . ,, ,.x

104 105 106 107 108 109

FIG. 4

To discuss the asymptotic behaviour of Hj(x), we start with a classical result: If vjx(x) is the number of all positive integers < x of the form (2.1), but without

the condition pi 1 (mod 4), then for each n = 1,2,3,... we have

1 x(loglog x) (2

-rn(X) -(n - 1)! log x ' ??) (2.5)

See [9, Ch. III, ?4, Th. 3.12] or [4, Sec. 22.18, Th. 437] (the case n = 1 is the prime number theorem).




Because (2.4) holds, we have to introduce a factor 2 for each prime factor of the form 4m + 1. From (2.5) it follows that

1 1 x(loglogx) ( ) (2.6)

Furthermore, for each n = 2,3,4,...,

Hn() 2(n -1) Hn-(X) 2(nlog1 (x oo), (2.7) Hn(x) log log x

and, therefore,

lim H(x) (2.8)

The asymptotic behaviour is completely different from what we have seen in TABLE 2 and FIGURE 4. However, x has to be "huge." The monotonic function on the right-hand side of (2.7) decreases extremely slowly. Its value is 1 at x = ee2( .

REMARK. In [8] and [15], the asymptotic behaviour of the number of all ppt with h < x is given by

00 x

2n-1 . H(x) 2 n=1

FORMULA OF LEGENDRE. For given h' = yh > 0, where h is of the form (2.1) and ,y does not contain divisors of the form 4m + 1, the total number M of Pythagorean triples that are corresponding to the hypotenuse h' is given by

2M + 1 = (2,81 + 1) - (2,82 + 1) * * * (213n + 1). (2.9)

This is a straightforward result from Theorem 1. We count all possible ppt, whose hypotenuses are divisors d of h'. The ppt can then be multiplied by h'/d to get the desired triples:

For n = 1, we have f81 triples: all possibilities are d = pk with 1 < k < f1. For n = 2, we have 2,11,2 + /1 + 12 triples, because d = pipk with j, k _ 1

has two common ppt. By induction, we arrive at the general case.

SOME CONSEQUENCES.

(1) A Pythagorean triple exists if and only if the hypotenuse has at least one prime factor of the form 4m + 1. Example: The representation for 10= 32 + 1 is unique.

(2) If y > 1, every triple corresponding to the hypotenuse h' is not primitive. (3) Every triple with the corresponding hypotenuse h' is primitive if and only if

h= P1 P2 ... pn with pairwise different primes pi 1 (mod 4). (4) The special case n = 1 in (2.9) implies that M= =31: For an arbitrarily

chosen integer M > 0 there is an infinity of hypotenuses h', each of them corresponding to M different Pythagorean triples. Example: The minimal value for h with three triples is 53. The triples are

52. [4,3,5] 5- [24,7,52] [336,527,53].



512 ALBERT FXSSLER [June-July

3. Triples with Common Leg-Sum s. From (1.2), for a given leg-sum we get

s = x + y = (a2 - b2) + 2ab = (a + b )2-2b2.

With u = a + b and v = b, this leads to Pell's equation

u2 - 2v2 = s with gcd(u, v) = 1. (3.1)

Because a > b, we have a + b > 2b and

u > 2v > O (3.2)

must also hold. Therefore, the solutions of (3.1) with the property (3.2) give the solutions of all

ppt with leg-sum s, where a = u - v, b = v. To deal with Pell's equation, we write

u2- 2V2 = (u + vC/ )(u - vC2 (3.3)

and we work in the integer ring Z[V2I consisting of elements a = u + vV2- with norm N(a) = U2 - 2V2. With the conjugate element a' = u - vC2, we obtain

N(a) = a - a' = u2-2v2, (3.4)

which is again (3.3). Indeed, solving Pell's equation is the same as finding elements with a given norm s.

Let us multiply the right-hand side of

(r2 - 2S2)(U2 - 2V2) = (r + s/2)(r - sx/)(u + vv2)(u - v2)

in two different ways: (1) first with third, and second with fourth factor; (2) first with fourth, and second with third factor.

Then we obtain

f (r2 - 2S2)(U2 - 2V2) = (ru + 2sv)2 -2 (rv + SU)2 (3.5) (r2 - 2S2)(U2 - 2V2) = (ru - 2sv)2 - 2 * (rv - su)

Elements with norm +1 or -1, so called units, play a significant role. The

complete set of units in Z[v2] is the following: +(1 + C2)m, m E Z. Their norms are equal to (- 1)W, because N(a * ,B) = N(a) N(,B) always holds. If we know one solution (uo, vo) of Pell's equation, we can generate recursively an infinite set of solutions by introducing the two units (1 + 4/_)2 = 3 + 2v24 and (3 + 2/2)-1 - 3 - 2a2 with norm + 1 in (3.5):

( Uk+1 + Vk+1 2 = (3 + 212)(uk + v/) = (3uk + 4vk) + ( +2uk + 3vk)v2

Uk1 + VkV1 2 = (3- 2V7)(Uk + v/2i) = (3Uk - 4Vk) + (-2uk + 3Vk)V2 (3.6)

k = 0,1,2,... Note that the indices in the first case are increasing and positive, in the second

case decreasing and negative. Two elements a and p in Z[VI are called associates if and only if a = ,B - c for

some unit E. This is an equivalence relation. All integers uj + vj/C2 generated in (3.6) form a subset of an equivalence class of Z[C2 ]: See [5, Ch. 16, Section 16.5].




The generalized identities

( (r2 + Rs2)(u2 + Rv2) = (ru - RsV)2 + R (rv + su)2

(r2 + RsS2)(U2 + RV2) = (ru + RsV)2 + R (rv -su)2

contain the two special cases (2.2) and (3.5) if we choose R= 1 and R= -2

respectively. In the previous section, we had the ring 2Z[i] of the Gaussian integers with norm N(u + iv) = lu + ijl = u2 + v2, where + 1, + i were the only units. The fact that we have an infinity of units makes the leg-sum problem tougher than the hypotenuse problem.

Now we use results from the Gaussian Law of Quadratic Reciprocity and

Quadratic Forms. See [9, Ch. VI, Th. 111], [11, Ch. 2], [1, Ch. 4].

LEMMA 1.

(A) There is a solution of problem (3.1) if and only if all prime factors of s are of the form 8m + 1.

(B) If s > 1 is a product of n different prime powers, each prime being of the form 8m ? 1, then among the infinity of solutions, there are exactly 2 n-1 pairwise nonassociated solutions of problem (3.1) with u > 0 and v > 0.

Note that 2n-1 is half the number of solutions of the equation U2 - 2 0 (mod s).

The question now is: How many of the infinity of solutions of (3.1) satisfy u > 2v > 0?

From

u2-4u2 2s-u2 = u - 2v > 0 (3.8)

u + 2v u + 2v

it follows that there are only a finite number of such pairs (u, v), if any. Trivially, if (u, v) is a solution of (3.1), then all four combinations (?u, ? v) are solutions. More is said in

LEMMA 2. Assumption: s has the form mentioned in (B). Among the infinity

(u, V) = (?u,, ? vk) of solutions of (3.1), where the set (Uk, Vk), k E Z is described in (3.6), there is exactly one which satisfies the condition u > 2v > 0.

Lemma 2 is proved in [12]. This article deals with the "Problemes de Fermat," in which Frenicle was involved too. See also [18, Ch. II, ?XII].

GENERATING PRIMITIVE PYTHAGOREAN TRIPLES WITH COMMON LEG-SUM. Us-

ing (3.5) for the n different prime powers of s > 1 gives 2 n-1 solutions of (3.1), if all prime factors of s are of the form 8m + 1. Iterating, in the sense of (3.6) until

luI > 21vI holds, generates the desired ppt with a = luI - IvI and b = Iv1.

Examples: The smallest leg-sum with two different ppt is s = 7- 17 = 119. u2 - 2V2= 7 has the solution u = 3, v = 1, and u2 - 2V2 = 17 has the solution u = 5, v = 2. Therefore, using (3.5), u2 - 2V2 = 119 has the solutions (u, v) = (19, 11) and (11, 1). The first does not satisfy the condition u > 2v. From the second line of (3.6) we have




One step is sufficient here. With (11, 1) and (13,5), we obtain the following result: [10, 1] generates [20,99, 101] and [8,5] generates [80,39,89].

The smallest leg-sum with 4 different ppt is s = 7 17 * 23 = 2737. Using

the two results for u, v above and the fact, that u2 - 2 v2 = 23 has the solution (11, 7) in each case after at most one iteration we get the generators: [37,36] [41, 16] [43, 12] [47, 6] with the ppt

[2664,73, 2737][1312, 1425, 2737][1032, 1705, 2737][564,2173,2737].

See also FIGURE 2, where the particular points are marked with s, assuming they appear in the diagram. There are more correspondences to the section dealing

with common hypotenuses:

(1) If on the right-hand side of (3.1) the leg-sum s > 1 and the prime factors

are of the form 8m + 1, then Theorem 1 holds for common s instead of common h, and so does the Legendre formula (2.9).

(2) Asymptotic behavior: Using r(x, 8, ? 1) in (2.3), the corresponding definition of Hn(x) (see definition following TABLE 1), denoted by Sn(x) and using the argument following (2.5), we get:

Sn-1 4(n -1) n=234 .............( Sn~ log log x

Sni lim = 0 (3.10) X noo

similar to (2.7) and (2.8).

4. Triples with Common Leg-difference, Area, Perimeter, Inradius or Leg.

1. Common Leg-difference d. With a common leg-difference d = Ix - y I and (1.3) we have

d = 12ab - (a2- b2)1

= 12a2 2 (a + b)21 with a > b > 0,

gcd(a, b) = 1, a + b = odd. (4.1)

Also in this case, we have to deal with problem (3.1), however without the restrictive condition (3.2). So it is not astonishing to get an infinity of solutions, if all prime factors of d are of the form 8 ? 1 and no solution otherwise. In [20] we can find the easily verifiable statement:

If [ a, b] generates a ppt with leg-diffeence d then so does [a + 2b, b].

Therefore, if we know one of the ppt, then an infinity of ppt can be generated.

2. Common Area f Searching for ppt with a common area f leads to the

problem of finding f such that the equation

1 f -* xy = ab(a-b)(a + b) with a > b > 0,

2

gcd(a, b) = 1, a + b = odd (4.2)

has several solutions. This problem is hard to solve. In [16], we learn that




The smallest area with a total of

(a) 2 ppt is 2 - 3 5 7 = 210, generated by [6,1][5,2]. (b) 3 ppt is 210 11 13 19 * 23 = 13123110, generated by

[77,38][78,55][138,5].

In [19], an infinity of pairs of ppt with common area, but no information on more than 2 ppt with common area is given. Despite Euler's efforts on this problem, not much more can be said today. The asymptotic behaviour for the number TA(x) of all ppt with area < x, denoted by TA(x), is given by (see [19]):

TA(x) c - ix with c = y 5/2 = 0.531340. (4.3)

We denote by Ak(x) the number of those areas < x which are associated with

exactly k ppt. With the aid of the computer, we found for x = 1,588,020: Al = 1976, A2= 22, A3 =1 and A4=A5 = ... = 0.

In [6] and [14], an efficient method for generating rational right triangles whose areas are integers is given.

3. Common Perimeter p

p = x + y + h = 2ab + (a2-b2) + (a2 + b2) = 2(a + b)a (4.4)

with the conditions mentioned in (4.2). Therefore p = even is the perimeter of a ppt if and only if

p ==u v with u>v>0 and gcd(u,v)=1. (4.5)

The generators are a = u - v, b = v. u > 2v and (4.5) imply the number of isoperimetrical ppt with even perimeter p is equal to the number of odd divisors u of p/2 with p7/ < u < IF and gcd(u, p/2v) = 1.

Using Bertrand's postulate, it is proven in [7], that for each m > 0, there is a perimeter p with at least m isoperimetrical ppt.

In [2], a list of 5 perimeters between 14,280 and 103,740 is given (60 < a < 210), each one associated with exactly 3 ppt. The asymptotic behaviour for the number Tp of all ppt with perimeter < x is given by (see [8]):

log 2 TA(x) 2 2 . (4.6)

4. Common Inradius i

2f ab(a + b)(a -b) i =b- = b(a -b) with a-b = odd. (4.7)

p (a +b)a

As a result of (4.7), i is always an integer, and the problem is simpler than that of the perimeter: If i has the prime factorization

i = 20p *p02 ... Po, with 2 <p1 p2< <pn

,8?0, f3 >1, (4.8)

then

the number of ppt with common inradius i is 2',




because it is equal to the number of ways to express i as a product of two relatively prime factors. An extensive study of this case can be found in [17, Sec. 23.2].

5. Common Leg. This problem is left to the reader.

5. Triples with Two Common Items.

THEOREM 3. There exists no two noncongruent right triangles with two of the following items in common: hypotenuse, leg-sum, leg-difference, area, perimeter, inradius, one leg.

Note that the theorem is not restricted to integers. Its result is quite obvious: 3 elements of a triangle essentially (with exceptions) determine it.

Proof. Some of the (I) possibilities are obvious, e.g., the ones with the common

leg. In any case, the proof is of the kind shown in the given representative examples below.

(a) Common hypotenuse h and inradius i. We use f for the area, p for the perimeter, and a for the angle where 0 < a < ir/4. In what follows, s stands for sin a and c for cos a.

We discuss the function

2f 2h2sc i(a) = i()=p =h + hs + hc

with a fixed value h.

1 di d sc c2 + c3 - s2 _ -3

h dae da 1 + s + c (+ s + c)2

(C-S)(C2 + CS + s2 + C + S)

(1 + s+c)

is clearly > 0 Va E (0, irr/4).

(b) Common leg-difference d = Ix - y I and inradius i. We consider 8 x - y to be fixed, then

i(x) = 2x(x + 8) 2x + 8 + X2 + (X + )2

With h = Vx2 + (X + 8)2 and D = 2x + 8 + h,

di [2 h 8] 5+8) D x- = [2(x + 5) + 2x][2x + 8 + h] -2 + 2x(x + ).

With y = x + 8 we have

D x- = (2x + 2y)(x + y + h) 2 + 2 xy dx h

= 2h3 + 2(x + y)(h2 - xy) > 0

because h2 _ xy > 0 for each right triangle.




Acknowledgements. I would like to thank my students Rudolf Eicher and Ramon Furer for their

computer work as well as for their experimental approach to number theory. Both stimulated me in this

work. Thanks also to L. A. Szekeley in Budapest for a private communication. Finally, I would like to

mention Prof. Robert Prielipp at the University of Wisconsin Oshkosh, Dr. Baoswan Wong-Dzung in

Untersiggenthal, Switzerland and Deynise Griggs in Columbia, South Carolina, who edited my text.

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Multiple Pythagorean Number Triples

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