Diophantus and Diophantine Equations (1)
( or so to say Special Equations)
( 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 17th
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....)