The first curve is the unit circle, which has many rational points, each corresponding to a Pythagorean Triple. To check (1,0) we substitute
[tex]x^2+y^2-1 = 1^2-0^2-1 = 0 \quad\checkmark[/tex]
That's a rational point, albeit a boring one. The first Pythagorean Triple on the list gives a more interesting point, (x,y)=(3/5, 4/5), which is also on the curve.
[tex] x^2 + y^2 - 3 = 0[/tex]
Let's assume there are rational points; we write them with a common denominator. Let x=a/d, y=b/d
[tex] (a/d)^2 + (b/d)^2 = 3[/tex]
[tex] a^2 + b^2 = 3d^2[/tex]
There are three cases:
Case 1: 3 divides exactly one of a or b
Case 2: 3 divides both a and b
Case 3: 3 divides neither a nor b
In case 1, let's say without loss of generality that it's a that's a multiple of 3. So is [tex]a^2[/tex] of course. But if b is not a multiple of 3, i.e [tex]b = 3n \pm 1[/tex] for some natural n, its square isn't a multiple of three either:
[tex](3n \pm 1)^2 = 3n^2 \pm 6n + 1 = 3(n^2 \pm 2n) + 1[/tex]
We see the square of a non-multiple of 3 necessarily has a remainder of 1 when divided by three.
So the sum of squares isn't a multiple of 3 so can't be [tex]3d^2[/tex]. That shows case 1 can't be the case.
In case 2, 3 divides both a and b, so each of the squares will be a power of 9 times some other factors, i.e. each square has an even number of factors of 3. So will the sum, because the other non-multiple of three factors when squared and added won't be a multiple of 3 as we showed in case 1. So d will have half of those factors of 3 and since there were an even number there won't be any leftover for the outside factor of 3. So case 2 can't be true.
Case 3. We showed both squares have remainder 1 when divided by 3, so their sum will have remainder 2, so can't be a multiple of 3.
None of the possibilities can give rise to rational points so we conclude there are no rational points on the curve.
