We say that a point p = (x, y) in r2 is rational if both x and y are rational. more precisely, p is rational if p = (x, y) ∈ q2. an equation f(x, y) = 0 is said to have a rational point if there exists x0, y0 ∈ q such that f(x0, y0) = 0. for example, the curve x2+y2−1=0 has rational point (x0,y0)=(1,0). show that the curve x2 + y2 − 3 = 0 has no rational points

The first curve is the unit circle, which has many rational points, each corresponding to a Pythagorean Triple. To check (1,0) we substitute

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.

Let's assume there are rational points; we write them with a common denominator. Let x=a/d, y=b/d

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 of course. But if b is not a multiple of 3, i.e for some natural n, its square isn't a multiple of three either:

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 . 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.

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