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The given equation is 30*x^2 +517 =  3*x^2y^2 +y^2

It is also given that x,y are integers.

The given equation can be rewritten as  

y^2*( 3x^2 +1) =  30*x^2 +517 or 

y^2 =   {30*x^2 +517} / ( 3x^2 +1).Dividing ,we get 

y^2 =  10+ (507)/ ( 3x^2 +1)

Let z = (507)/ ( 3x^2 +1).Now as y is an integer, y^2 is  also  an integer.  Further as   

y^2 =  10+z ,   the value of  z should be such that y^2= z+10 is a perfect square,so that y is an integer.Now (507)/ ( 3x^2 +1) =  3*13*13/  ( 3x^2 +1), where 3,13,13  are the prime factors of 507.Therefore the value of the expression 3x^2 +1 should be a factor  of the numerator 507,so that z becomes an integer when divided by it.The factors of 507 are 1,3,13,39,169 and 507. That is the expression 3x^2+1, can take values 

1,3,13,39,169&507 only. The corresponding values of z are 507,169,39,13,1. Of these z=39 is the only possible value because z+10 =39+10 =49 is a perfect square only when z=39.Other factors will not result in a perfect square value for y^2. Hence y^2 =49

Now z = 39 = 507/ 3x^2+1  or  3x^2+1 =   507/39 = 13 or x^2 =  (13-1)/3 =  4 .So x^2 = 4.

Hence x^2+y^2 =4+49 =53

 

 

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