Diophantine Equations (Special Equations)

Diophantus and Diophantine Equations (1)
( or so to say Special Equations)

Diophantus an Alexandrian Greek Mathematician was born sometime between AD 201 and 215 and died aged 84 sometime between AD 285 and 299. He is often considered as the father of algebra.

While studying Diophantus' Arithmetica, Pierre de Fermat concluded in 17^th century that a certain equation considered by Diophantus had no solutions; and noted in the margin without elaboration that he had found "a truly marvelous proof of this proposition," now referred to as Fermat's Last Theorem. This led to remarkable advances in Number Theory, and the study of Diophantine Equations, Diophantine Geometry and Diophantine Approximations. In modern use, Diophantine equations are usually algebraic equations with integer coefficients, for which integer solutions are sought. 

These Diophantine equations, or   so to say, Special equations, are a great source of Mathematical Recreation. Moreover such equations do appear in many avatars in various Competitive exams.

For example one encounters such questions: Find the no. of integral solutions or Positive integral solutions of the following equations:

(1)    ax +  by = c

(2)    1/x  + 1/y  = 1/n

(3)    1/x – 1/y = 1/n

(4)    a/x + b/y  = 1/n

(5)    1/x²  + 1/y²  = 1/n²

(6)    a² + b² =c²

(7)    aⁿ + bⁿ =cⁿ

(8)    x² – ny² = ±1

Solution to the above equations will be posted in several blog posts in the coming weeks.

Let us discuss in this post about the Positive Integral solution of the following Diophantine equation: 

  xyz = 4( x + y + z)

We can write the above equation as below:

xyz/ ( x + y + z) =4

In order to solve the above equation let us assume x ≥ y≥ z

Case 1: Let us take the minimum possible value of z=1.

xy/(x + y=1) = 4 

Now putting y= 1, 2,3,4,5, 6 etc we find that

When y=1 ; x is negative.

When y=2 ; x is negative.

When y=3 ; x is negative

When y=4 ; x can not be found.

When y=5; x is equal to 24

When y=6; x is equal to 14

When y=7; x is equal to 32/3 which is not integral and hence not acceptable

When y=8; x is equal to 9

When y=9; x is equal to 8. Clearly this violates the condition x ≥ y≥ z. For all the values above y=9 this violation will take place.

Hence the no of positive integral solution in Case 1 is equal to only 3 (when x ≥ y≥ z)

(i)                 z=1, y=5, x = 24

(ii)               z=1, y=6, x = 14

(iii)             z=1, y=8, x = 9

Case 2:  Let us take z=2.

xy/(x + y=1) = 4

Now putting y= 1, 2,3,4,5, 6 etc we find that

When y=1; x is negative.

When y=2; x can not be found.

When y=3; x =10

When y=4; x =6

When y=5; x is equal to 14/3 which is not integral.

When y=6 x is equal to 1. Clearly this violates the condition x ≥ y≥ z. For all the values above y=6 this violation will take place.

Hence only two solutions are possible in this case:

(i)                 z=2  y=3 x= 10

(ii)               z=2  y=4  x=6

For z = 3, 4, 5, 6 there are no solutions.

Hence no of integral solutions = 5 when x ≥ y≥ z

But as x, y and z may be in different ascending or descending order,  

The total no solutions = 5x 3! = 30

                                                                                            ( To be Continued....)