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A set $S = leftlbrace left( x, y right) vert x^2 + y^2 = dfrac1n^2, text where n in mathbbN text and either x in mathbbQ text or y in mathbbQ rightrbrace$ is given. I need to prove that this is countable.

I have tried looking for a bijection $f: mathbbQ rightarrow S$ as $f left( x right) = left( x, y right)$ where $y$ is a fixed real number such that the property of the set is satisfied.

Clearly, this function is not even a surjection.

How should we define our bijection so that we prove $S$ is countable?

If you index on $x$ and $n$, you can do the following. Let
$$
E_n=mathbb Qcap left[-tfrac1n^2,tfrac1n^2right].
$$
This sets $E_n$ are countable, because they are subsets of $mathbb Q$.
Then $S=S_1cup S_2$, where
$$
S_1=bigcup_ninmathbb Nbigcup_xin E_nleftleft(x,sqrttfrac1n^2-x^2right)rightcupleftleft(x,-sqrttfrac1n^2-x^2right)right
$$
$$
S_2=bigcup_ninmathbb Nbigcup_yin E_nleftleft(sqrttfrac1n^2-y^2,yright)rightcupleftleft(-sqrttfrac1n^2-y^2,yright)right
$$
We can map this to a subset of $(mathbb Ntimesmathbb Qtimes1,2)^2$. And this last set is countable, so $S$ is countable.

The key fact is that subsets of countable sets are countable, and that finite cartesian products of countable sets are countable.

Note that $x^2=frac1n^2-y^2$ and $y^2=frac1n^2-x^2$. So if $y$ is rational, then $x^2$ is rational as well, and likewise for $y^2$ if $x$ is rational.

Therefore you can define a surjection from the collection of all roots of rational numbers (which also includes all the rationals). So now comes the question, can you prove that this set is countable?