Edit: {The answer to your question, "...do we already know that a positive binary form represents arbitrarily long arithmetic progressions?" is yes. See the second paragraph below.}
If the relative density exists, so does the Dirichlet density and they are equal. The converse is not true in general. For primes in a given arithmetic progression, both densities exist. See Lang's Algebraic Number Theory, Ch. VIII.4 and XV. Given those facts, one approach to the problem would be trying to show that the relative density of the set of primes represented by a positive binary quadratic form actually exists (I have no idea how hard this might be).
On the other hand, if you only want to know about a.p.'s of primes represented by a positive quadratic form, a better approach might be answering the question, "Does Green-Tao still hold for sets of primes with positive Dirichlet density?" The answer is yes since G-T only requires that the limsup of the relative density be positive, and positive Dirichlet density implies positive limsup (if the limsup were 0, the lim would be 0).
There's a short-cut in Roth's approach if one only cares to get $o(N)$. Adolf Hildebrand told me so, and here is my shortest writeup.
Notation: Let $r(N),\rho(N)$ be the largest cardinality and density of a subset of $[N]$ that is free of 3-term APs, and let $\rho=\lim \rho(N)$, which must exist by Fekete's subadditive lemma since $r(N+M)\leq r(N)+r(M)$. Let $A\subset [N]$ witness $r(N)$, and let $S(x) = \sum_{a\in A} e( a x)$ (where $e(ax)=\exp(2\pi i a x)$, naturally). Let $T(x)=\sum_{n=1}^N e(nx)$.
Lemmata: Now $r(N)=|A|=I:=\int_0^1 S(x)^2 S(-2x)dx$, since $A$ is 3-free. By Parseval, $|A|=\int_0^1 |S(x)|^2dx$. By expanding $S(x)^2 T(-2x)$ and exchanging integration and summation, $(|A|/2)^2 \leq I_0:=\int_0^1 S(x)^2 T(-2x)$.
Main Engine: As long as $0< M\leq N$, $$\sup_{x\in {\mathbb R}} |S(x)-\rho(M) T(x)| \leq N(\rho(M)-\rho(N)) + 10 M \sqrt{N}.$$ Proof is by circle method; set $E(x):=S(x)-\rho(M)T(x)$. Three steps are
- $|E(a/q)|\leq N(\rho(M)-\rho(N))+2Mq$, the hard one;
- $|E(a/q+\beta)| \leq N(\rho(M)-\rho(N))+2Mq +2\pi|\beta|NMq$;
- $|E(x)| \leq N(\rho(M)-\rho(N))+10M\sqrt{N}$.
Main Lemma: By Main Engine and Lemmata, $\Delta:=|I-\rho(M)I_0|$ is at most $(N(\rho(M)-\rho(N))+10M\sqrt{N})|A|$. By Lemmata, $\Delta$ is at least $|A|(\rho(M)\rho(N)N/4-1)$. Therefore, $$\rho(N)\rho(M)\leq 4(\rho(M)-\rho(N))+50M/\sqrt{N}.$$
Conclusion: Let $N\to\infty$ and then $M\to\infty$ to get $\rho^2\leq 0$.
The step labeled "the hard one" in the Main Engine is tricky, and uses the fact that we can restrict $A$ to arithmetic progressions and still get a 3-free set. The length of the progressions we restrict to is connected to $M$.
I suppose the point is that a write-up can be almost arbitrarily short or extremely long, depending on what the reader can be trusted to fill in.
Best Answer
"The Discovery of Incommensurability" by Kurt von Fritz [ http://www.jstor.org/stable/1969021 ] indicates that the early Greek mathematicians did not explicitly use the Fundamental Theorem to prove the irrationality of √2. The proof known to Aristotle ("the diagonal of the square is incommensurate with the side, because odd numbers are equal to evens if it is supposed to be commensurate") uses a restricted version of the Fundamental Theorem, as explained in http://en.wikipedia.org/wiki/Quadratic_irrational
Apparently, the explicit use of the Fundamental Theorem to prove the irrationality of √2 is post-Gauss. This is argued convincingly by Barry Mazur:
This fundamental theorem of arithmetic has a peculiar history. It is not trivial, and any of its proofs take work, and, indeed, are interesting in themselves. But it is nowhere stated in the ancient literature. It was used, implicitly, by the early modern mathematicians, Euler included, without anyone noticing that it actually required some verification, until Gauss finally realized the need for stating it explicitly, and proving it.