Proof for a circle with x^2 + y^2 = 3?

Question

Is there only exist irrational coordinates ( as in one of either x or y is irational) to a circle of radius 3? if so what is the proof for it.

Answer 1

If there are rational numbers then choose integral a, b, c, d such that

(a/b)^2 + (c/d)^2 = 3 and a/b and c/d are in lowest terms. Then there is an integral solution a^2 + c^2 = 3b^2d^2.

The right side is divisible by 3, so the left side must be too. Look at a and c mod 3. The squares mod 3 are 0, 1, 1 for 0, 1, 2. For the left side to be divisible by 3, the only way to do this is for both a and c to be divisible by 3. (otherwise the left would be congruent to 1 or 2 mod 3). This means both a and c are divisible by 3, so a=3x and b = 3y for some x and y, and so a^2 + b^2 = 9x^2 + 9y^2 = 3b^2d^2. The left is divisible by 9, so the right is too.

This means that 3 divides b or 3 divides d - but this contradicts the selection of a/b and c/d in lowest terms, so there is no such a, b, c, d, so there is no rational solution.

(sorry about the earlier incorrect proof)

Answer 2

This isn't a circle of radius 3, this is a circle of radius √3. Different things, esp. since a circle of radius 3 contains several points with rational coordinates, such as (0, 3).

Suppose that (x, y) is on this circle and suppose that x∈Q, and y∈Q. Then x=p/q for some coprime integers p and q, and likewise y=r/s for some coprime integers r and s. Then:

x²+y²=3
p²/q² + r²/s² = 3
p²s² + r²q² = 3q²s²

Now from this we obtain that:

p²s² = (3s²-r²)q²

So q² divides p²s². However, since p and q are coprime, so too are p² and q², thus by Euclid's lemma, q² divides s². However, we can also obtain:

r²q² = (3q²-p²)s²

So s² divides r²q², and since s and r are coprime, so too are s² and r², so invoking Euclid's lemma again, we find that s² divides q². But s² and q² are both natural numbers, so s²|q² and q²|s² implies q²=s². So we have:

p²/q² + r²/s² = (p²+r²)/q² = 3
p²+r²=3q²

Now, for any integer n, n²≡0 mod 4 or n²≡1 mod 4. Suppose q²≡1 mod 4. Then 3q²≡3 mod 4. However, p²≡0 or p²≡1 mod 4, and likewise r²≡0 or r²≡1 mod 4, so p² + r² ≡ 0, 1, or 2 mod 4, and in particular, p²+r²≠3 mod 4 (that symbol should be "not congruent to", not "not equal to", but I'm not sure whether the "not congruent to" symbol (≢) will show up on your system). So it is impossible that q²≡1 mod 4, thus q²≡0 mod 4. However, if p²≡1 mod 4, then p²+r²≡1 or 2 mod 4, neither of which is 0. So p²≡0 mod 4. However, this implies that both p and q are multiples of 2, contradicting the assumption that they are coprime.

So it is impossible that x²+y²=3, with both x and y rational. Q.E.D.

Edit: nevermind, sofarsogood answer is simpler than mine.

Answer 3

Yes, there are only irrational coordinates. First, x and y have to be less than the square root of three which is about 1.7, so the only possible integer either could be is 1. There is no Pythagorean triplet with 1 with a hypotenuse of 3.

I don't know how detailed you want this to be... if you wanted to you could prove that the square root of 3 is irrational by contradiction.