Fractional parts.

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

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

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

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

$\overline{CP_{\lfloor{\dfrac{n}{2}\rfloor + 1}}} = (n - (\lfloor{\dfrac{n}{2}}\rfloor + 1) + \dfrac{1}{2} + \dfrac{1}{\sqrt{2}})x =$

$\dfrac{2(n - \lfloor{\dfrac{n}{2}}\rfloor) + \sqrt{2} - 1}{2}x$

Let $a_{n} = n - \lfloor{\dfrac{n}{2}}\rfloor \implies$ $\overline{CP_{\lfloor{\dfrac{n}{2}\rfloor + 1}}} = \dfrac{\sqrt{2}(2a_{n} + \sqrt{2} - 1)a}{3\sqrt{2} + 2n - 1}$

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

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

Let $b_{n} = \lfloor{\dfrac{n}{2}\rfloor}$

$\implies A_{\triangle{2}} = (\overline{AP_{b_{n}}})^2 = \dfrac{8(b_{n} + \sqrt{2} - 1)^2}{(2n + 3\sqrt{2} - 1)^2}a^2$

$\implies A_{\triangle{1}} + A_{\triangle{2}} = \dfrac{2(4(a_{n}^2 + b_{n}^2) + 4(\sqrt{2} - 1)(a_{n} + 2b_{n}) + 5(3 - \sqrt{2}))}{(2n + 3\sqrt{2} - 1)^2}a^2$

$a_{n} = n - \lfloor{\dfrac{n}{2}}\rfloor = \dfrac{n}{2} + \left\{\dfrac{n}{2}\right\}$ and $b_{n} = \lfloor{\dfrac{n}{2}\rfloor} = \dfrac{n}{2} - \left\{\dfrac{n}{2}\right\}$

$\implies A_{\triangle{1}} + A_{\triangle{2}} =$ $\dfrac{2(2n^2 + 6(\sqrt{2} - 1)n + 5(3 - 2\sqrt{2}) + 8(\left\{\dfrac{n}{2}\right\})^2 - 4\left\{\dfrac{n}{2}\right\}}{(2n + 3\sqrt{2} - 1)^2}a^2$

$A_{T} = a^2 - (A_{\triangle{1}} + A_{\triangle{2}})$

$\implies \dfrac{A_{T}}{A_{ABCD}} = \dfrac{8n + 14\sqrt{2} - 11}{(2n + 3\sqrt{2} - 1)^2} + \dfrac{8(\sqrt{2} - 1)\left\{\dfrac{n}{2}\right\} - 16(\left\{\dfrac{n}{2}\right\})^2}{(2n + 3\sqrt{2} - 1)^2}$

$= \dfrac{8(\sqrt{2} - 1)\left\{\dfrac{n}{2}\right\} - 16(\left\{\dfrac{n}{2}\right\})^2}{(2n + 3\sqrt{2} - 1)^2} + \dfrac{212\sqrt{2} - 281}{49} \implies$

$\dfrac{8n + 14\sqrt{2} - 11}{(2n + 3\sqrt{2} - 1)^2} = \dfrac{212\sqrt{2} - 281}{49} \implies$

$392n + 686\sqrt{2} - 539 = (848\sqrt{2} - 1124)n^2 + (6212 - 4220\sqrt{2})n + 5714\sqrt{2} - 7883$

$\implies$

$(212\sqrt{2} - 281)n^2 + (1455 - 1055\sqrt{2})n + 1257\sqrt{2} - 1836 = 0$

After simplifying and dropping the negative root we obtain:

$n = \dfrac{(212\sqrt{2} + 281)\sqrt{147539 - 100254\sqrt{2}} + 38465 - 12005\sqrt{2}}{21854}$

$= \dfrac{(212\sqrt{2} + 218)(\dfrac{7(6913\sqrt{2} - 7359)}{223}) + 38465 - 12005\sqrt{2}}{21854} =$

$\dfrac{2677115\sqrt{2} + 6042631 + 8577695 - 2677115\sqrt{2}}{4873442} = \dfrac{14620326}{4873442} = \boxed{3}$ .