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Tuesday, September 16, 2008

Q) Find the number of positive integral solutions of x4 _ y4 = 3789108 .

A)
Let A = x^2 and B = y^2
A^2 - B^2 = 3789108
(A + B)(A - B) = 3789108

Let A + B = T
(T)(T-2B) = 3789108
T^2 - 2BT - 3789108 = 0
T = 2B +- root (4B^2 + 4*3789108) / 2
T = B +- root (B^2 + 3789108)
A + B = B +- root (B^2 + 3789108)
A = +- root (B^2 + 3789108)
Now since A = x^2 so A cannot be negative, thus
A = root (B^2 + 3789108)
Now A is x^2, and B is y^2, where x and y are integers.
Squares of integers (ie. x^2 and y^2) end in 0, 1, 4, 5, 9, 6 and squares of these perfect squares (ie. A^2 and B^2) end in 0, 1, 5,6.
Now,
last digit of B^2 + last digit of 3789108 should end in 0,1,5,6
the various sum of last digits possible are 8, 9, 3, 4 none of which is 0, 1, 5 or 6...
So no integral solution exists.

The equation has 0 integral solutions.


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1 comment:

Anonymous said...

Buddy,
A^2-B^2=3789108
so,A^2=3789108+B^2
A=sqrt(3789108+B^2)

Now since A = x^2 so A cannot be negative, thus
A = root (B^2 + 3789108)
Now A is x^2, and B is y^2, where x and y are integers.
Squares of integers (ie. x^2 and y^2) end in 0, 1, 4, 5, 9, 6 and squares of these perfect squares (ie. A^2 and B^2) end in 0, 1, 5,6.
Now,
last digit of B^2 + last digit of 3789108 should end in 0,1,5,6
the various sum of last digits possible are 8, 9, 3, 4 none of which is 0, 1, 5 or 6...
So no integral solution exists.

The equation has 0 integral solutions.