More of Circles and Squares.

Let $n$ be a fixed even integer, $a$ a side of square $ABCD$ and $x$ a side of square $AJP_{1}I$ .

Extending the diagram to even $n$ congruent circles we obtain:

$\sqrt{2}a = \sqrt{2}x + (n - 1)x + \dfrac{x}{2} + \dfrac{x}{\sqrt{2}} \implies 4a = (6 + \sqrt{2}(2n - 1))x \implies$

$x = \dfrac{4a}{6 + \sqrt{2}(2n - 1)} = \dfrac{4a}{\sqrt{2}(3\sqrt{2} + (2n - 1))} =$ $\dfrac{2\sqrt{2}}{3\sqrt{2} + (2n - 1)}a$

$\overline{CP_{\frac{n}{2} + 1}} = (n - (\dfrac{n}{2} + 1) + \dfrac{1}{2} + \dfrac{1}{\sqrt{2}})x =$ $(\dfrac{\sqrt{2}n + 2 - \sqrt{2}}{2\sqrt{2}})(\dfrac{2\sqrt{2}}{2n + 3\sqrt{2} - 1})a =$

$\dfrac{\sqrt{2}(n + \sqrt{2} - 1)}{2n + 3\sqrt{2} - 1}a \implies$ $A_{\triangle{1}} = (\overline{CP_{\frac{n}{2} + 1}})^2 =$ $\dfrac{2(n + \sqrt{2} - 1)^2}{(2n + 3\sqrt{2} - 1)^2}a^2$

$\overline{AP_{\frac{n}{2}}} = (\sqrt{2} + (\dfrac{n}{2} - 1))x = (\dfrac{n + 2(\sqrt{2} - 1)}{2})(\dfrac{2\sqrt{2}}{2n + 3\sqrt{2} - 1})a =$

$\dfrac{\sqrt{2}(n + 2(\sqrt{2} - 1))}{2n + 3\sqrt{2} - 1}a \implies$ $A_{\triangle{2}} = (\overline{AP_{\frac{n}{2}}})^2 = \dfrac{2(n + 2(\sqrt{2} - 1))^2}{(2n + 3\sqrt{2} - 1)^2}a^2$

$\implies A_{\triangle{1}} + A_{\triangle{2}} = \dfrac{2(2n^2 + 6(\sqrt{2} - 1)n + 5(3 - 2\sqrt{2}))}{(2n + 3\sqrt{2} - 1)^2} a^2 \implies A_{T} = a^2 - (A_{\triangle{1}} + A_{\triangle{2}}) \implies$

$\dfrac{A_{T}}{A_{\triangle{ABCD}}} = \dfrac{8n + 14\sqrt{2} - 11}{(2n + 3\sqrt{2} - 1)^2} = \dfrac{32\sqrt{2} - 41}{9}$

$\implies (128\sqrt{2} - 164)n^2 + (860 - 620\sqrt{2})n + 728\sqrt{2} - 1064 = 0 \implies$

$n = \dfrac{-(860 - 620\sqrt{2}) \pm (204 - 108\sqrt{2})}{2(128\sqrt{2} - 164)}$

For $(+): n = \dfrac{-656 + 512\sqrt{2}}{2(128\sqrt{2} - 164)} =$ $\dfrac{4(128\sqrt{2} - 164)}{2(128\sqrt{2} - 164)} = 2$

For $(-): n = \dfrac{371 - 525\sqrt{2}}{367} < 0 \therefore$ dropping negative irrational root $\implies \boxed{n = 2}$ .