take 3*x^2+y^2-4*y-17 =a and 2*x^2+2*y^2-4*y-6 =b then a^3-b^3=(a-b)^3 we know a^3-b^3=(a-b)^3-3*a*b*(a-b) so a=0,b=0,a-b=0 x^2-y^2-11=0 (x+y)(x-y)=11 x=6 and y=5
Following the same procedure as you have suggested, we get the solutions a=0 or b=0 or a=b For the case a=b, we get x^2-y^2-11=0 ie.,(x+y)(x-y)=11. As 11 can be written as 11., 1.11, -1.-11, -11.-1 there will be four possibilities. further a=0 gives 3x^2+y^2-4y-17=0, which gives 3x^2+(y-2)^2=21. This will give x=2 or -2 and y=5 or -1 (four possibilities) Further b=0 gives 2x^2+2y^2-4y-6=0 ie., x^2+(y-1)^2=4 this gives the possibilities x=2 or -2 and y=1 OR x=0 and y=3, or -1 again four possibilities. So, in total there are 12 solutions
, determine its solutions (x, y) where both x and y are integers. 
3 comments:
take 3*x^2+y^2-4*y-17 =a
and 2*x^2+2*y^2-4*y-6 =b
then a^3-b^3=(a-b)^3
we know a^3-b^3=(a-b)^3-3*a*b*(a-b)
so a=0,b=0,a-b=0
x^2-y^2-11=0
(x+y)(x-y)=11
x=6 and y=5
The Catch is at the step "So a=0,b=0,a-b=0"
Is it "or" in between them or "and"?
That is what is going to decide the solutions
Following the same procedure as you have suggested, we get the solutions a=0 or b=0 or a=b
For the case a=b, we get x^2-y^2-11=0
ie.,(x+y)(x-y)=11.
As 11 can be written as 11., 1.11, -1.-11, -11.-1 there will be four possibilities. further a=0 gives 3x^2+y^2-4y-17=0, which gives 3x^2+(y-2)^2=21. This will give x=2 or -2 and y=5 or -1 (four possibilities)
Further b=0 gives 2x^2+2y^2-4y-6=0 ie., x^2+(y-1)^2=4
this gives the possibilities x=2 or -2 and y=1 OR x=0 and y=3, or -1 again four possibilities. So, in total there are 12 solutions
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