Int. J. of Mathematical Sciences and Applications,
Vol. 1, No. 3, September 2011
Copyright  Mind Reader Publications
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SPECIAL PYTHAGOREAN TRIANGLES WITH PERIMETER AS A SUM
OF TWO SQUARES AND A CUBE
Mita Darbari
Head, Deptt. Of Mathematics, St. Aloysius College, Jabalpur, India
e-mail: m.darbari@rediffmail.com
ABSTRACT
Special Pythagorean Triangles, in terms of perimeter to be sum of two squares and a
cube, are obtained. Various 3D graphs of corresponding Pythagorean triplets are plotted A special
case when two sides are consecutive is obtained. A few interesting results are observed.
KEY WORDS: Pythagorean Triangles, sum of two legs is a cube; Mathematica
INTRODUCTION:
In the Pythagorean Mathematics, the determination of integral solutions of cubic Diophantine equations by
Gopalan and Premlatha [1], Janaki and Gopalan [2], Gopalan, Somnath and Vanitha[3], Pandichelvi [4], Gopalan and
Pandichelvi [5], Gopalan and Vijyasankar [6], Gopalan and Srividhya [7] and classifying all Pythagorean triangles
wherein each of which the perimeter can always be represented as the sum of three distinct squares by Gopalan and
Devibala [8], motivated us to examine whether there exists any Pythagorean Triangle with sum of its legs a cube. It is
towards this end we make an attempt to find patterns of Special Pythagorean Triangles with perimeter as sum of two
squares and a cube. A special case when two sides are consecutive is also obtained.
METHOD OF ANALYSIS: The primitive solutions of the Pythagorean Equation
X2 + Y2 = Z2
(1)
is given by [9]
X = m2 - n2, Y = 2mn, Z = m2 + n2
(2)
for some integers m, n of opposite parity such that m > n > 0 and (m, n) = 1.
I Sum of two legs is a cube: X + Y = β3
which gives,
m2 - n2 + 2mn = β3
(3)
Then solutions of (3) is given by
β =( m2 - n2 + 2mn)1/3,
β =-(-1)1/3 (m2 - n2 + 2mn)1/3,
β= (-1)2/3 (m2 - n2 + 2mn)1/3
(4)
II Perimeter is sum of two squares and a cube:
From (2) and (3),
X + Y + Z = β3 + m2 + n2
(5)
(2) and (4) generates X and Y which satisfy (1) and (3) in correspondence with (5).
Few examples are given in the Table 1 below:
TABLE 1
1
2
3
4
5
6
7
C
3
m
n
X
Y
Z
X
+
Y
=
β
β
R
1
14
7
147
196
245
147 + 196 = 343 = 73
7
2
16
247
96
265
247 + 96 = 343 = 73
7
3
3
38
1976
2120
768 + 1976 = 2744 = 143
14
26 768
4
42
1176
1960
1568 + 1176 = 2744 = 143
14
14 1568
5
51
3468
3757
1445 + 3468 = 4913 = 173
17
34 1445
6
63
3905
1008
4033
3905 + 1008 = 4913 = 173
17
8
7
78
12012
12013
155 + 12012 = 12167 = 233
23
77 155
8
92
4232
8993
7935 + 4232 = 12167 = 233
23
23 7935
9
112
12544
15680
9408 + 12544 = 21952 = 283
28
56 9408
10
15808 + 6144 = 21952 = 283
128
24 15808
6144
16960
28
1225
Mita Darbari
1.
2.
3.
We observe that
For every value of β, there are two pairs of m and n.
As given in row 10, with β given, m and n can be found out
Considering the special case as in row 1, 4, 8, n = β, we get equation (3) as m2 - n2 + 2mn = n3, which gives
m = -n-√ (2 n2 + n3) and m = -n+√( 2 n2 + n3).
As √ (2 n2 + n3) > n, discarding the first value, we get
m = -n + √ (2 n2 + n3)
(6)
Putting n = β = p2 – 2,
where p is a positive integer greater than 2.
(7)
From (3.2.6) and (3.2.7) we get
m = (p2 – 2) (p– 1)
(8)
Therefore, (3.2.2) gives
X = (p2 – 2)2 (p2 – 2p),
Y = 2(p2 – 2) 2 (p– 1),
Z = (p2 – 2)2 (p2 – 2p + 2)
(9)
X + Y = (p2 – 2)3 = β3
From (2), (7) and (8),
X + Y + Z = (p2 – 2)3 + (p2 – 2)2(p– 1) 2 + (p2 – 2)2
(10)
(9) generates X, Y and Z which satisfy (1) and (3) in correspondence with (7) and (10).
Few examples are given in the table 2 below:
TABLE 2
1
2
3
6
7
8
9
C
3
p
m
n
Z
β
X+Y=β
X + Y + Z = β3 + m2 + n2
R
3
1
3
14
7
245
147 + 196 = 7
588 = 73 + 142 + 72
7
3
2
4
42
14
1960
1568 + 1176 = 14
4704 = 143 + 422 + 142
14
3
3
5
92
23
8993
7935 + 4232 = 23
21160 = 233 + 922 + 232
23
4
6
170
34
30056
27744 + 11560= 39304 69360 = 343 + 1702 + 342
34
= 343
5
7
282
47
81733
47
77315 + 26508 =
185556 = 473 + 2822 + 432
3
103823 = 47
6
8
434
62
192200
62
184512 + 53816 =
430528 = 623 + 4342 + 622
3
238328= 62
7
9
632
79
405665
79
393183 + 100488 =
899336 = 793 + 6322 + 792
3
493671 = 79
8
10 882
98
787528
98
768320+ 172872 =
1728720 = 983 + 8822 + 982
3
941192 = 98
9
11 1190 119 1430261
119
1401939 + 283220 =
3115420 = 1193 + 11902 + 1192
3
1685159 = 119
10
142
2419680 + 443608 =
5323296 = 1423 + 15622 + 1422
12 1562 142 2460008
3
2863288 = 142
III One leg and a side are consecutive:
In such cases, m = n + 1, then (3) gives 2n2+4n+1- β3 == 0, using software Mathematica for 0 < n < 100000000000 &
0 < β < 10000, there is just one solution for n = 77, β = 23 ! Therefore,
m = 78, which with equations (2), gives
X = 155, Y = 12012, Z = 12013 and X2 + Y2 = Z2 , i.e., 1552 + 120122 = 120132
, i.e., 24025 + 144288144 = 144312169 = 120132
For n ≥ 1010, the existence of special Pythagorean triangles with two sides consecutive and perimeter to be sum of two
squares and a cube, needs to be investigated.
Solving the Diophantine equation m2 + 2mn + n2 - β3 = 0 when n < m && 0< m< 105 && 0 < n < 105 && 0 < β < 105,
we get 1993 solutions. Some of the special Pythagorean Triplets (X, Y, Z) with sum of legs to be a cube and perimeter
to be sum of two squares and a cube are listed below with values less than 107:
(147,196,245),(247,96,265),(768,1976,2120),(1568,1176,1960),(1445,3468,3757),(3905,1008,4033),(155,12012,1201
3),(7935,4232,8993),(9408,12544,15680),(15808,6144,16960),(6727,23064,24025),(17751,12040,21449),(8064,3124
0,32264),(27744,11560,30056),(31937,36984,48865),(35301,33620,48749),(33856,63480,71944),(96096,1240,96104
1226
SPECIAL PYTHAGOREAN TRIANGLES WITH PERIMETER…
),(33495,70328,77897),(77315,26508,81733),(21609,96040,98441),(57229,60420,83221),(77469,40180,87269),(4915
2,126464,135680),(100352,75264,125440),(96320,142008,171592),(184512,53816,192200),(107163,142884,178605)
,(180063,69984,193185),(92480,221952,240448),(249920,64512,258112),(55451,302460,307501),(148211,209700,2
56789),(48793,340224,343705),(239805,149212,282437),(229419,263620,349469),(393183,99856,405665),(268960,
282408,389992),(295872,255496,390920),(261393,443576,514865),(650093,54876,652405),(9920,768768,768832),(
507840,270848,575552),(212064,618520,653864),(562624,267960,623176),(122317,790356,799765),(330105,58256
8,669593),(321440,619752,698152),(483360,457832,665768),(768320,172872,787528),(434283,658444,788765),(58
3495,509232,774457).
3D PLOTS: Plotting these 1993 solutions we get following 3D graphs:
1.5 1010
1.0 1010
5.0 109
1.5 1010
0
0
1.0 1010
5.0 109
5 109
1 1010
0
Figure 1: ListPointPlot3D
Figure 2: ListPlot3D
Figure 3: ListSurfacePlot3D
1.
2.
3.
4.
OBSERVATIONS
We also observe that
X + Y + Z = 0(mod 2)
( Y + Z – X )2 = 2( Y + Z)(Z – X)
( X + 2Y + Z )2 = (Z – X )2 + 4( X + Y)(Y + Z)
X + 2Y + Z ± 2{ ( X + Y)(Z - Y)}1/2 = 0(mod16) or = 0(mod4)
In conclusion, one may attempt to find other patterns of Pythagorean Triangle which satisfy the conditions presented
in the above problem.
REFERENCES
1.
M.A. Gopalan, S. Premlatha; On the Diophantine Equation with Five Unknowns x3 + y3 + w(x2 – y2 ) = 2z3 + (x +
y)p2; Antarctica J. Math.; 7(2); 2010; Pages:213-218.
2.
G. Janaki, M.A. Gopalan; Integral Soutions of x2 – y2 + xy = (m2 – 5n2) z3; Antarctica J. Math.; 7(1); 2010; Pages:6367.
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Mita Darbari
3.
M.A. Gopalan, Somnath Manju, N. Vanitha; Integral Soutions of Ternary Cubic Equation x3 + y3 + 4z2 = 3xy (x + y);
Antarctica J. Math.;7(3); 2010; Pages:311-315.
4.
V. Pandichelvi; Observations on the ternary cubic equation xy + x + y + 1 = (2a +1)z3; Antarctica J. Math.; 7(3);
2010; Pages:273-278.
5.
M.A. Gopalan, V. Pandichelvi; Remarkable Soutions on the cubic equation with four unknowns x3 + y3 + z3 = 28 (x +
y + z)w2; Antarctica J. Math.; 7(4); 2010; P.393- 401.
6.
M.A. Gopalan, A. Vijyasankar; Integral Soutions of Ternary Equation x2 + y2 – xy + 2(x + y + 2) = z3; Antarctica J.
Math.; 7(4); 2010; Pages:455-460.
7.
M.A. Gopalan , G. Srividhya Krishnamoorthy; ON TERNARY CUBIC DIOPHANTINE EQUATION ; Global Journal
of Pure and applied Mathematics (Vol.4 No.3); 2008.
M.A. Gopalan, S.Devibala; Special Pythagorean Triangles; The Mathematics Education,Vol. XLI(4); Siwan Printers;
Siwan; Dec 2007; Page No. 293
Burton David M.; Elementary Number Theory; Tata McGraw Hill; New Delhi; 2009; Page No. 248.
8.
9.
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