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Algebra
Comments (7)
@ Hari Shankar
The question says only positive integers
@Rahul
a very good answer indeed.
But the result can be obtained with much less effort and in a more general way
case 1. a=b
in this case u get a =b = sqrt (3) ; hence no soln
case 2. a!=b
sice the expression is symmetric wrt a and b , without any loss of generality , we may assume a > b
then we get
1 < 3/b^2
or b< sqrt(3)
the only possibility is then b = 1
which gives 1/a^2 + 1/a = 0 , which gives no solution for positive integer 'a' as the LHS > 0
(proved)
Another way is to look at the expression as a quadratic in one of the variables. WLOG let it be a.
Then we have the quadratic a2(b2-1) -ab-b2=0 whose discriminant is b2(4b2-3) which should be a perfect square which happens when b=1 or -1. Correspondingly, a=-1 and a=1
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a=1, b=-1 and a=-1, b=1