Prove that x^2-y^2=2 has no integral solutions.

Ok lets see, write the equation as:

x^2 = y^2 + 2

Now, so

x = (y^2 + 2)^(1/2)

Now the above equation represents a hyperbola ( by general form: (x^2)/(a^2) - (y^2)/(b^2) =1 )

so, it will be symmetric,

so if it have any integral coordiantes ( or any integral solution of above equation in any of the four quadrants then it should have in all four , and if not in any of one then in none of them )

so lets take out the first quadrant,

here x and y, both are positive,

Now vary the value of y so that we can get corresponding values of x (and can see if both be integers)

lets start by least integer (considering the first quadrant) which is zero, 0

so if y = 0,

x=(2)^(1/2)=1.414 , not an integer

now let y = 1, so x=(3)^(1/2) = 1.732,not an integer again

now let y=2, so x=(4 + 2)^(1/2) = 6^(1/2), again not an integer

Now from 2 and above, the perfect squares will be at a distance more than 2, so cannot get a perfect square by adding a perfect square with 2

REASON: we have to get:    Perfect square = another Perfect square + 2

Now for 2, square = 2*2

for 3 square = 3*3,

so the distance between consecutive square increases as the number increases,

so no integral solutions can be obtained afterwards,

Best of Luck for your exams

(x+y)(x-y)=2

2 is a prime no.

hence either x+y=2 x-y=1

               or vice versa

in either case not posible

Let us assume that x^2-y^2=2 has a integral solution.

Then (x+y)(x-y)=2

the only way 2 can be factorized is 2 = 2*1

As x+y>x-y we can assume that x+y=2 ....................... 1 and

                                              x-y=1.........................2

From 1   x=2-y  and from 2   x = 1+y

therefore y is an integer between 1 and 2.....................this is a contradiction to our assumption.

Hence our assumption must be wrong.

Hence x^2-y^2=2 has no integral solution.  [HENCE PROVED].

Please approve my solution in case you like it.

                                                                  ADHIRAJ MANDAL