Does every strip between two parallel lines of positive irrational slope contain a point with perfect square coordinates? Equivalently (I think), are there perfect square points arbitrarily close to any positively irrationally sloped line?
EDIT: [wrong, retracted].
EDIT 2: Numerical experiments seem to hint at a counterexample for $\alpha = \frac{3-\sqrt{5}}{2} = (\phi -1)^2$. With $p_k/q_k$ being convergents of the continued fraction $[0;\overline{1}]$ for $\phi-1$, it seems that
$$\lim\limits_{k \to \infty}{\left|p_k^2-\alpha q_k^2\right|} = 0.55278\dots$$
so square points other than $(0,0)$ are at least that constant away vertically from $y = \alpha x$. Squared perpendicular distance from the points to the line seems to tend to a rational, $4/15$.
EDIT 3: The convergents in EDIT 2 are the ratios of consecutive Fibonacci numbers $F_n$. With the help of wxMaxima I managed to calculate:
$$\lim_{n\to \infty}\left| {{\left(\phi-1\right)^2\ F_{n+1}^2}}-F_n^2\right| = \sqrt{\frac{6-2\sqrt{5}}{5}} ≈ 0.55278... > 0$$
which settles the general question negatively, and I'm nearly sure that a similar limit exists whenever the CF for $\sqrt{\alpha}$ is eventually periodic (i.e. $\alpha$ is a square of a quadratic irrational) — or more precisely, the smallest of a finite set of limits, one for each offset within the period.
As the answer is geting somewhat messy, here's a summary:
For all $\alpha>0,\beta\in\mathbb R,n>1$ and $\epsilon>0$, is there $p,q\in\mathbb N$ such that
$$\left|q^n\alpha+\beta-p^n\right|<\epsilon$$
For $n<2$, the answer is yes. For $n>2$, then we can find many counterexamples. In the case of $n=2$, it depends how well can $\sqrt\alpha$ be rationally approximated.
Edit 2
The edit here is just the discussion in the comments properly written out + more generality:
Let $\alpha$ be any irrational, and let $\frac{p_n}{q_n}$ be the continued fractions of $\alpha$. We know that they satisfy the following minimal property:
For all $0<q'\leq q_n$, we have $$\left|q_n\alpha-p_n\right|\leq\left|q'\alpha-p'\right|$$ and equality only holds when $q'=q_n,p'=p_n$. Note that this is one way to define the covergents of $\alpha$.
Define $$\varepsilon_q=\min_{0<p,0<q'\leq q}\left|q\alpha-p\right|$$
we note that the values of $\varepsilon_q$ are precisely those that comes from the continued fraction.
Suppose $\alpha,p,q>0$ and $\left|q^2\alpha^2-p^2\right|<\varepsilon$. We now have the following chain of inequalities (for small enough $\varepsilon$, i.e. $\varepsilon<q^2\alpha^2$):
\begin{align*} \epsilon&>\left|q^2\alpha^2-p^2\right|\\ &=\left|q\alpha-p\right|\left|q\alpha+p\right|\\ &\geq\varepsilon_q\left(q\alpha+\sqrt{q^2\alpha^2-\varepsilon}\right)\\ &=\varepsilon_qq\alpha\left(1+\sqrt{1-\frac{\varepsilon}{q^2\alpha^2}}\right)\\ &\geq\varepsilon_qq\alpha\\ \end{align*}
and by choosing $\alpha=\frac{1+\sqrt5}2$, we know that the convergents must be given by $\frac{F_{n+1}}{F_n}$ where $F_n$ is the Fibbonaci sequence. With some effort, we can show that
\begin{align*} \varepsilon_{F_n}&=\left|F_n\alpha-F_{n+1}\right|\\ \varepsilon_{F_n}F_n&=\left|\frac1{\sqrt5}\left((-1)^n-\left(\frac{1-\sqrt5}2\right)^{2n}\right)\right|\\ &\geq\frac{3-\sqrt5}2\approx0.382 \end{align*}
which approaches $\frac1{\sqrt5}$ as $n\to\infty$. This shows that
$$\left|q^2\alpha^2-p^2\right|\geq\left(\frac32-\frac{\sqrt5}2\right)\alpha=\frac{1-\sqrt5}{2}$$ .
Hence
$$\left|q^2\alpha^2+\beta-p^2\right|\geq\frac{1-\sqrt5}{2}-|\beta|$$
and if we choose any $|\beta|<\frac{1-\sqrt5}{2}$, then the quantity $\left|q^2\alpha^2+\beta-p^2\right|$ will never get arbitrarily close to $0$ for any $p,q\in\mathbb Z$
Note: By requiring $\varepsilon_qq$ doesn't eventually reach $0$, this is slightly stronger than asking for the irrationality measure to be $2$ (all irrational algebraic integers have irrationality measure $2$ - see Roth's theorem). It may be interesting to study what happens when the irrationality measure is above $2$, or even say $\alpha$ is some Liouville number. The bounding would certainly be much harder but it could be doable? Another thing that this technique can't work on is when $|\beta|$ is reasonably big which may be interesting to look at as well.
Note that we can also bound in the other direction to see what numbers `works':
\begin{align*} &\left|q_n\alpha-p_n\right|\left|q_n\alpha+p_n\right|\\ \leq&\varepsilon_{q_n}\left(2q_n\alpha+\varepsilon_{q_n}\right)\\ =&2q_n\varepsilon_{q_n}\alpha+\varepsilon_{q_n}^2\\ \end{align*}
and if $\lim_{n\to\infty}q_n\varepsilon_{q_n}=0$, we see that $\left|q_n^2\alpha^2-p_n^2\right|\to0$ as well. A classical result (that isn't too hard to prove) is that the sign of $q_n\alpha-p_n$ is $\pm(-1)^n$, which directly tells us that for any $\beta,\epsilon$, we can find some $p,q\in\mathbb Z$ such that $\left|q^2\alpha^2+\beta-p^2\right|<\epsilon$ (Hint: Split the case for $\beta>0$ and $\beta<0$, and multiply $q^2\alpha^2+\beta-p^2$ by $n^2$ for an appropriate $n$).
From the discussion in the previous paragraph, $\lim_{n\to\infty}\varepsilon_{q_n}q_n>0\iff\exists\beta$ where we cannot approximate $\left|q^2\alpha^2+\beta-p^2\right|$ arbitrarily well.
Previous attempt - solution for exponent $>2$ with some 'size' bounds
Not quite the answer but hopefully it can be modified to work, so far this only works if the exponent is above $2$:
For some strip given by $\alpha x+\beta_1,\alpha x+\beta_2$, if we want to find some $(q^2,p^2)$ in it, it is equivalent to asking for $\left|\alpha q^2+\frac{\beta_1+\beta_2}2-p^2\right|<\frac{\left|\beta_1-\beta_2\right|}2$, hence it is sufficient to show if for all $\alpha,\beta,\epsilon$, do there exists $p,q$
$$\left|\alpha q^2-\beta-p^2\right|<\epsilon$$
However we want to omit the trivial cases which occurs where $\alpha,\beta$ are rational, hence a better question is perhaps for 'almost all $\alpha,\beta$'. This suggests the following approach:
Choose some $f$ such that $\int_{-\infty}^\infty f=1$.
Fix $p,q,\epsilon$, we find the measure of all possible $\alpha,\beta$:
$$\int_{-\infty}^\infty\int_{\alpha q^2-p^2-\epsilon}^{\alpha q^2-p^2+\epsilon}fd\beta fd\alpha$$
then if for some $\epsilon$ we have
$$\sum_{p=1}^\infty\sum_{q=1}^\infty\int_{-\infty}^\infty\int_{\alpha q^2-p^2-\epsilon}^{\alpha q^2-p^2+\epsilon}fd\beta fd\alpha<1$$ we would be done.
The hope is that since $p^2,q^2$ 'grows quite quickly', the sum converges.
For instance, say we choose $f(x)=\frac12\text{sech}^2(x)$, we get
\begin{align*} \int_{\alpha q^2-p^2-\epsilon}^{\alpha q^2-p^2+\epsilon}fd\beta&=\frac12\tanh\left(\alpha q^2-p^2+\epsilon\right)-\frac12\tanh\left(\alpha q^2-p^2-\epsilon\right)\\ &=\frac{\sinh(2\epsilon)}{\cosh\left(2(p^2-q^2\alpha)\right)+\cosh(2\epsilon)} \end{align*}
and this gives us the following sum:
\begin{align*} &\sum_{p=1}^\infty\sum_{q=1}^\infty\int_{-\infty}^\infty\int_{\alpha q^2-p^2-\epsilon}^{\alpha q^2-p^2+\epsilon}fd\beta fd\alpha\\ =&\sum_{p=1}^\infty\sum_{q=1}^\infty\int_{-\infty}^\infty\frac12\frac{\text{sech}^2(\alpha)\sinh(2\epsilon)}{\cosh\left(2(p^2-q^2\alpha)\right)+\cosh(2\epsilon)}d\alpha\\ =&\frac12\sinh(2\epsilon)\sum_{p=1}^\infty\sum_{q=1}^\infty\int_{-\infty}^\infty\frac{\text{sech}^2(\alpha)}{\cosh\left(2(p^2-q^2\alpha)\right)+\cosh(2\epsilon)}d\alpha\\ <&\frac12\sinh(2\epsilon)\sum_{p=1}^\infty\sum_{q=1}^\infty\int_{-\infty}^\infty\frac{\text{sech}^2(\alpha)}{\cosh\left(2(p^2-q^2\alpha)\right)+1}d\alpha\\ =&\frac14\sinh(2\epsilon)\sum_{p=1}^\infty\sum_{q=1}^\infty\int_{-\infty}^\infty\text{sech}^2(\alpha)\text{sech}^2(p^2-q^2\alpha)d\alpha\\ \end{align*}
Unfortunately, the $\sum_{p=1}^\infty$ seems to numerically be about $\frac1q$, which means the sum diverges and we need a better choice of $f$.
Interestingly, if we use $(q^3,p^3)$ instead, numerical evidence shows that the sum grows like $\frac1{q^2}$ which does converge, so this method may still be viable.
If all else fails, we could still obtain interesting results by setting say $\epsilon$ as a function of $p$ or $q$ which gives us some bounds on what is possible and what isn't.
Edit
With a bit more playing around, I think I got a clearer sketch in this case. This shows that the result holds for $(q^{2+\epsilon},p^{2+\epsilon})$ for $\epsilon>0$, not quite what we want but close.
Let $n=2+\epsilon$ as above, our goal now is to show the following term is finite:
$$\sum_{p=1}^\infty\sum_{q=1}^\infty\int_{-\infty}^\infty\text{sech}^2(\alpha)\text{sech}^2(p^n-q^n\alpha)d\alpha$$
We note that $\text{sech}^2(x)<e^{-2|x|}$. This simplfies our integral down to
\begin{align*} &\int_{-\infty}^\infty\text{sech}^2(\alpha)\text{sech}^2(p^n-q^n\alpha)d\alpha\\ <&\int_{-\infty}^\infty e^{-2|\alpha|}e^{-2\left|p^n-q^n\alpha\right|}d\alpha\\ =&\frac{e^{-2\left(\frac pq\right)^nq^n}-e^{-2p^n}}{2q^{-2n}-2} \end{align*}
When summing over $p,q$, the $$\frac{-e^{-2p^n}}{2q^{-2n}-2}$$ term converges. The other term can be shown to converge with the following bound:
\begin{align*} &\sum_{p=1}^\infty e^{-2\left(\frac pq\right)^n}\\ <&\int_0^\infty e^{-2\left(\frac pq\right)^n}dp\\ =&2^{-\frac1n}q^{-1}\Gamma\left(1+\frac1n\right)\\ \end{align*}
and we're done since $$\sum_{q=1}^\infty\frac{q^{n-1}}{2q^{-2n}-2}$$ converges for $n>2$.
To summarize, this tells us that if we take any $n>2$, we can find $\epsilon$ small enough such that there exists some positive measure of $\alpha,\beta$ where the following inequality is never true for any $p,q\in\mathbb Z$:
$$\left|\alpha q^n-\beta-p^n\right|<\epsilon$$