4. A SIMPLE SOLUTION FOR DIOPHANTINE EQUATIONS OF SECOND, THIRD

AND FOURTH POWER [Journal No. “Mapana”- MJS. Vol.4, No.1, June.2005, Published by, Christ College, Bangalore]

SCOPE OF RESEARCH

We know already that the set of positive integers, which are satisfying the Pythagoras equation of three

variables and four variables are called Pythagorean triples & quadruples respectively. These are

Diophantine equation of second power. The all unknowns in this Pythagorean equation have already been

solved by mathematicians Euclid & Diophantine. However the solution defined by Euclid & Diophantine

is also again having unknowns. The only possible to solve the Diophantine equations was trial & error

method. Moreover, the trial & error method to obtain these values are not so practical and easy especially

for time bound works, since the Diophantine equations are having more than two unknown variables. The

scope of work is to solve the (1) Pythagorean triples1, (2) Pythagorean Quadruples2 & n-tuples and (3)

Diophantine equations of third & fourth power3 by simple method. After conducting various exercises,

the author has realized that there is a mathematical relation in between the variables and he has attempted

to establish the necessary formulae to solve these equations. These formulae & methods have been proved

with appropriate examples. It is very useful for Students, Research scholars and persons those who are

preparing question papers related to right-angled triangles, rectangular prisms and Number theory4.

THE NEW CONCEPT

The new concept, which is now developed by the author is that the mathematical relation in between the

variables in a Diophantine equation is that: If a Diophantine equation

x1m + x2

m + x3m + x4

m + … … + xnm + ym = Zmis written in the form X + ym = Zm,

one of the pair of the conjugate pair factors of X is the key factor to slove the equation. where, X = x1

m + x2m + x3

m + x4m + … . . +xn

m..

EXISTING METHODS:

For Diophantine equation of Pythagorean triples 𝑥2 + 𝑦2 = 𝑟2, the solution defined by mathematicians

Euclid & Diophantine as “All Pythagorean triples in which ‘x’, ‘y’, ‘r’ are without common factor and ‘x’

is odd are obtained by replacing the letters ‘a’ and ‘b’ in the triple {(𝑎2 − 𝑏2), (2𝑎𝑏), (𝑎2 + 𝑏2)} by

whole numbers that have odd sum and no common factor”. For Diophantine equation of Pythagorean

quadruples 𝑥2 + 𝑦2 + 𝑧2 = 𝑟2, the solution defined by mathematicians Euclid & Diophantine as “All

Pythagorean triples in which ‘x’, ‘y’, ‘r’ are without common factor and ‘x’ is odd are obtained by

replacing the letters ‘a’, ‘b’, ‘c’ & ‘d’ in the quadruple{(a2 − b2 + c2 − d2), (2bc + 2ad), (a2 + b2 +c2 + d2)} by whole numbers that have odd sum and no common factor”. These equations having further

unknown variables and it requires trial & error method. The Diophantine equation of n-tuples of second

power, can be solved easily by using the following formulae & the methods.

NEW METHODS TO SOLVE DIOPHANTINE EQUATIONS:

1. Procedure to solve the Diophantine equation of second power of n-tuples

The Diophantine equation of n-tuples of second power is 𝑥12 + 𝑥2

2 + 𝑥32 + 𝑥4

2 + … … . +𝑥𝑛2 + 𝑦2 = 𝑍2.

Let, (𝑥12 + 𝑥2

2 + 𝑥32 + 𝑥4

2 + … … . +𝑥𝑛2) = 𝑋

and substituting in the above equation, we get, X + y2 = Z2.

For given value of x1, let us assume any value for 𝑥2, 𝑥3, 𝑥4 … . . 𝑥𝑛 (if these values are not given). So we

have to find the value of y and z only. Identify all of the effective conjugate pair factors 𝑝 𝑎𝑛𝑑 𝑞of X,

but ‘p’ should be less than ‘q’ and also both p & q should be even or odd.

By substituting the effective set of values of 𝑝 𝑎𝑛𝑑 𝑞 in following formula, the value of ‘y’ and ‘z’ can

be determined.

(𝑦, 𝑧) = ([𝑞 − 𝑝

2] , [

𝑞 + 𝑝

2])

𝐓𝐡𝐞𝐫𝐞𝐟𝐨𝐫𝐞, 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐏𝐲𝐭𝐡𝐚𝐠𝐨𝐫𝐞𝐚𝐧 𝐧 − 𝐭𝐮𝐩𝐥𝐞𝐬 𝐚𝐫𝐞 {𝐱𝟏, 𝐱𝟐, 𝐱𝟑, 𝐱𝟒, … … . . 𝐱𝐧, 𝐲, 𝐳}

Result & Discussion:

Some results are given in Table-1

Table-1

Example

No.

x1 x2 x3 x4 x5 X Conjugate pair

factor

y z

p q

1 24 - - - - 576 6 96 45 51

2 35 - - - - 1225 7 175 84 91

3 5 4 - - - 41 1 41 20 21

4 4 8 - - - 80 2 40 19 21

5 2 4 5 - - 45 3 15 6 9

6 3 4 6 8 - 125 5 25 10 15

Practical Example-1

A man has to travel from A to B by a shortest route. There is a problem and he could not travel on the

road. He has traveled 33 km to reach B via C. The perpendicular distance of C from the road A to B is 12

km. What is the shortest distance from A to B?

Fig. 1.1

Answer:

The given data is AC + CB = 33 km and CD = 12km. CD is the common side for both right-triangles

ADC and BDC. Now we need to calculate the other sides and hypotenuse. The set of Pythagorean triples

for right-triangle whose one of its side is 12 are as follows;

122 = 144

The conjugate pair factors of 144 are (2 × 72), (4 × 36), (6 × 24), (8 × 18), The sides (x, y, r) are given below

(x, y, r) = (12,q − p

2,q + p

2)

= (12,72 − 2

2,72 + 2

2)

= (12, 35, 37) − − − − − − − − − − − − − − − − − − − − − − − − − − − (1.1)

(x, y, r) = (12,q − p

2,q + p

2)

= (12,36 − 4

2,36 + 4

2)

= (12, 16, 20) − − − − − − − − − − − − − − − − − − − − − − − − − − − (1.2)

(x, y, r) = (12,q − p

2,q + p

2)

= (16,24 − 6

2,24 + 6

2)

= (12, 9, 15) − − − − − − − − − − − − − − − − − − − − − − − − − − − −(1.3)

(x, y, r) = (12,q − p

2,q + p

2)

= (12,18 − 8

2,18 + 8

2)

= (12, 5, 13) − − − − − − − − − − − − − − − − − − − − − − − − − − − −(1.4)

Fig. 1.2

Fig. 1.3

The relevant two right angled triangles are (12, 16, 20) & (5, 12, 13) i.e. (from eqn. 1.2 & 1.3)

Fig. 1.4

Therefore, AD = 16 km, DC = 12 km, AC = 20 km, BD = 5 km, & BC = 13 km

The shortest distance of AB = AD + BD = 16 + 5 = 21 km

Practical Example-2

There is a road to travel from A to C via B. The shortest distance from A to C is 42 km. A man who want

to travel from A to C reached B. After arriving B, he come to know that there is a problem in the bridge

which is in between B and C and he cannot travel by the road from B to C. However, he took 90º left

diversion and travelled 16 km to arrive D. Then he arrived C by a shortest route from D. How much he

has travelled extra through D than the shortest route from A to C?

Fig. 2.1

Answer:

The given data is BD = 16 km and AB + BD + DC = 42 km. BD is the common side for both right-

triangles ABD and CBD. As one of the sides is 16, find out other sides and hypotenuse. The set of

Pythagorean triples for right-triangle whose one of its side is 16 are as follows;

162 = 256

The conjugate pair factors of 256 are (2 × 128), (4 × 64) and (8 × 32) The sides are

(x, y, r) = (16,q − p

2,q + p

2)

= (16,128 − 2

2,128 + 2

2)

= (16, 63, 65) − − − − − − − − − − − − − − − − − − − (2.1)

(𝑥, 𝑦, 𝑟) = (16,𝑞 − 𝑝

2,𝑞 + 𝑝

2)

= (16,64 − 4

2,64 + 4

2)

= (16, 30, 34) − − − − − − − − − − − − − − − − − − − (22)

(𝑥, 𝑦, 𝑟) = (16,𝑞 − 𝑝

2,𝑞 + 𝑝

2)

= (16,32 − 8

2,32 + 8

2)

= (16, 12, 20) − − − − − − − − − − − − − − − − − − − (2.3)

Fig. 2.2

Fig. 2.3

The combination of 12 and 30 makes the value 42 and corresponding r values are 20 and 34.

Fig. 2.4

Therefore, AB = 12; AD = 20; BC = 30 & DC = 34

The total travelled distance = 12 + 16 + 34 = 62 km

The shortest distance = 42 km

The extra distance travelled = 62 − 42 = 20 km.

Practical Example-3

A man has to travel from A to B. The road is blocked unexpectedly due to some repair work in between A

& B and he is unable travel on the road. He has traveled via C and D. The distance of BD is 15 km, which

is exactly perpendicular to the AC. While return to A from B, the repair work is in progress in between B

and D. He arrived at A via C and D. If the shortest distance from A to B is 148 km. What is the difference

in distance between onward and return travel?

Fig. 3.1

Answer:

The given data is CD = 15 km and AD + DB = 148 km. CD is the common side for both right-triangles

ADC and CDB. As one of the sides is 15, find out other sides and hypotenuse. The set of Pythagorean

triples for right-triangle whose one of its side is 15 are as follows;

152 = 225

The conjugate pair factors of 225 are (1 × 225), (3 × 75), (5 × 45) and (9 × 25) The sides are

(x, y, r) = (15,q − p

2,q + p

2)

= (15,225 − 1

2,225 + 1

2)

= (15, 112, 113) − − − − − − − − − − − − − − − − − − − (3.1)

(x, y, r) = (15,q − p

2,q + p

2)

= (15,75 − 3

2,75 + 3

2)

= (15, 36, 39) − − − − − − − − − − − − − − − − − − − (3.2)

(x, y, r) = (15,q − p

2,q + p

2)

= (15,45 − 5

2,45 + 5

2)

= (15, 20, 25) − − − − − − − − − − − − − − − − − − − (3.3)

(x, y, r) = (15,q − p

2,q + p

2)

= (15,25 − 9

2,25 + 9

2)

= (15, 8, 17) − − − − − − − − − − − − − − − − − − − (3.4)

Fig. 3.2

Fig. 3.3

AB = 148 km

The combination of y values 112 & 36 make the value 148 km and corresponding r values are 113 and 39.

Fig. 3.4

Fig. 3.5

Therefore, AD = 36; AC = 39; BD = 112 & BC = 113

The distance trvelled from A to B via C & D (onward) = AC + CD + DB = 39 + 15 + 112 = 166 km

The distance trvelled from B to A via C & D (return) = 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 = 113 + 15 + 36 = 164 km

Therefore, The difference between onward and return travel = 166 − 164 = 2 km.

2. Procedure to solve the Diophantine equation of third power of n-tuples

The Diophantine equation of third power of any n-tuples is

x13 + x2

3 + x33 + x4

3 + … … . +xn3 + y3 = Z3

Let,(x13 + x2

3 + x33 + x4

3 + … … . +xn3) = X and substituting in above equation, we get X + y3 = Z3 For

given value of 1x , assume any value for 𝑥2, 𝑥3, 𝑥4 … . . 𝑥𝑛 (if these values are not given), but 𝑛 ≥ 2. The

set of values of {𝑦, 𝑧} can be obtained from the following formula.

(𝑦, 𝑧) = (√3(4𝑞 − 𝑝2) − 3𝑝

6,

√3(4𝑞 − 𝑝2) + 3𝑝

6)

The value of √3(4q − p2) should be integer and ≰ 3p.

Where, p and q are the conjugate pair factor of ‘X’.

Result & discussion

Some examples & its results are given in Table-2

Table-2

Example

No.

x1 x2 x3 x4 x5 X Conjugate pair factor y z

p q

1 3 5 - - - 152 2 76 4 6

2 6 8 - - - 728 8 91 1 9

3 11 15 - - - 4706 2 2353 27 29

4 5 7 9 - - 1197 3 399 10 13

5 11 12 13 - - 5256 6 876 14 20

6 1 3 4 5 - 217 1 217 8 9

7 3 4 5 8 - 728 2 364 10 12

8 1 5 6 7 8 1197 3 399 10 13

Procedure to solve the Diophantine equation of fourth power of n-tuples

The Diophantine equation of third power of any n-tuples is

x14 + x2

4 + x34 + x4

4 + … … . . +xn4 + y4 = z4

Let, (x14 + x2

4 + x34 + x4

4 + … … . . +xn4 ) = X and substituting in above equation, we get X + y4 = z4.

For given value of 1x , assume any value for x2, x3 … . . xn ,2x (if these values are not given), but n ≥ 2

and xn ≠ 0. The set of values of (y, z) can be obtained from the following formula.

(𝑦, 𝑧) = (√(𝑞 − 𝑝

2) , √(

𝑞 + 𝑝

2))

Where, 𝐩 and 𝐪 are the conjugate pair factor of ‘X’

Result & discussion

The different examples & its results are given in Table-1:

Table-1

Example

No.

x1 x2 x3 x4 x5 X Conjugate pair

factor

y z

p q

1 2 2 3 4 - 369 9 41 4 5

2 4 6 8 14 - 44064 144 306 9 15

3 21 4 22 26 - 885969 441 2009 315 353

CONCLUSION

The new method to solve the Diophantine equation of n-tuples of second power, third power, fourth

power have been explained in this paper and which has been examined and proved in the chapter of

Result & discussion with appropriate examples. It is very useful for Students, Research scholars and etc.

Refer new glossary of mathematical terms in Appendix-1, which have been developed now by the author

APPENDIX-1:

Conjugate pair factors:

The factor and corresponding quotient of any number are called conjugate pair factor of that number.

e.g: If R = p × q, p and q are the conjugate pair factor of R.

REFERENCES

1. W.Gellert.S.Gottwald, M.Hellwich. H.Kästner.H.Küstner, The VNR Concise Encyclopedia of

Mathematics, Van Nostrand Reinhold, New York, 1989 p.672.

2. McGraw-Hill Encyclopedia of Science & Technology, Vol-14, 9th Edition, New York, 2002,

p.649.

3. http:// www.mathworld.wolfram.com

4. http:// www.mathworld.wolfram.com