^{4 _ }y

^{4}= 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:

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.

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