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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
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