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Assuming $a,b$ are integers, it can be verified by brute-force search that if there are any solutions, then $a\ge4000$ or $b\ge2000$. Checked with Mathematica with following code: Monitor[Do[If[Abs[Log[5]/Log[7] - a/b] <= 1/7^b, Print[{a, b}]], {b, 1, 1999}, {a, 1, 3999}], {b, a}] Using Will Jagy's advice, checking continued fractions up to 19 terms ...


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Probably not en-style with the rules, but after few days I came to this solution, which probably isn't the most elegant. Still looking for a better one... If $a$ is a perfect square itself, we're done. Now, let's assume ... $a$ is not a perfect square, then $\sqrt{a}\in \mathbb{R} \setminus \mathbb{Q}$ and so is $\frac{1}{\sqrt{a}}\in \mathbb{R} \setminus ...


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This is probably not what you want, but suppose you want a perfect square starting with $a_1a_2\dots a_n$ then for sufficiently large $2k$ we shall have $10^k\sqrt{a_1a_2\dots a_n+1}-10^k\sqrt{a_1a_2\dots a_n}= 10^k(\sqrt{a_1a_2\dots a_n+1}-\sqrt{a_1a_2\dots a_n})>1$. This means there is a perfect square between $10^{2k}(a_1a_2\dots a_n)$ and ...



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