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Examples of irrational real values x & y are:

  • $a \sqrt{c}$ & $b \sqrt{c}$ where a, b are rational integers with $a^2 + b^2 = c$, e.g. $\sqrt{5}$ & $2 \sqrt{5}$

  • $\sqrt{a^2 + b} - \sqrt{b}$ & $\sqrt{a^2 + b} + \sqrt{b}$, e.g. $\sqrt{7} - \sqrt{3}$ & $\sqrt{7} + \sqrt{3}$

If the requirement for $x^2 + y^2$ to be a perfect square is artificial or contrived then feel free to discuss the apparently more general case where it can be any rational integer, especially if this leads to a more natural and satisfactory classification.

Thinking about the problem, I would have guessed that cyclotomic integers might be involved, were it not for the fact that non-trivial cyclotomic integers are complex, whereas for this question $x$ & $y$ must both be real. But since $x^2 + y^2 = (x - i y)(x + i y)$ it may be that, as some famous mathematician once asserted, the route to the real domain lies through the complex domain!

(As people have remarked in other problems similar to this, I have an uneasy feeling I have overlooked some obvious observation that makes the problem trivial! )

John R Ramsden
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    What exactly is the problem? For any positive algebraic $c$ and real algebraic $x$ with $x^2 \le c$, $y = \pm\sqrt{c - x^2}$ is real and algebraic and $x^2 + y^2 = c$. – Robert Israel May 09 '19 at 16:03
  • Hmm, yes. It does seem I've made a big fuss over a problem which is indeed trivial! – John R Ramsden May 09 '19 at 17:23
  • To put this in context, I was thinking about the problem of finding a point in the plane of a square such that the point is at rational distances from all four corners of the square. So maybe in relation to that there is some restriction of the problem I asked which makes it more "interesting". – John R Ramsden May 09 '19 at 17:39

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You can parameterize the values.

$x^2+y^2=c$ Let

$x=\sqrt{c}\cdot \cos(\theta)$

$y=\sqrt{c}\cdot \sin(\theta)$

You get a suitable x and y for every theta.

If you want algebraic values, you'll need to restrict theta to angles of algebraic sines.

We have let $\sin(\theta)=u$ and find $\theta$ by taking the inverse sine of $u$.

Another approach to get some algebraic sines:

$\sin{(\theta/2)}=\sqrt{\frac{1-\cos{\theta}}{2}}$

So for a given angle with algebraic sine, you can generate infinite algebraic sines by cutting it in half, then iterate.

Found a related question:

Algebraic Sines question

TurlocTheRed
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  • Yes, that was the first thing I also considered. But somehow it seems like begging the question, i.e. simply shifting the burden to classifying "algebraic sines", But then if that is easy then I suppose it would answer the question. – John R Ramsden May 09 '19 at 15:17
  • I think you can prove you have a suitable x and y iff there is a corresponding theta. – TurlocTheRed May 09 '19 at 15:31