One fold of rectangle can know everything?

Let $Q$ , $R$ , $S$ , and $T$ be the vertices of the rectangle, $MN$ be the paper's crease with length $65$ , $P$ be the intersection of the crease and the paper's diagonal, and $MO$ be perpendicular to $ST$ , as shown below. Also let $a$ and $b$ be the lengths of the paper's two sides $QT$ and $ST$ respectively, and $c$ be the length of $NT$ .

Since $Q$ reflects on $S$ in crease $MN$ through $P$ , $QP \cong PS$ and $\angle SPN \cong \angle QPN$ . Since $PN \cong PN$ by the reflexive property, $\triangle SPN \cong \triangle QPN$ by SAS congruence. Therefore, $NQ = NS = b - c$ , and by Pythagorean's Theorem on $\triangle NTQ$ , $a^2 + c^2 = (b - c)^2$ . Solving this equation for $c$ gives $c = \frac{b^2 - a^2}{2b}$ .

By rotational symmetry, $RM = NT$ , and since $RMSO$ is a rectangle, $SO = RM$ . Therefore, $RM = SO = NT = c$ . Since $b = SO + ON + NT$ , $b = ON + 2c$ , and so $ON = b - 2c$ . Since $QMOT$ is a rectangle, $MO = QT = a$ . Therefore, by Pythagorean's Theorem on $\triangle MON$ , $a^2 + (b - 2c)^2 = 65^2$ .

Substituting $c = \frac{b^2 - a^2}{2b}$ into $a^2 + (b - 2c)^2 = 65^2$ gives $a^2 + \frac{a^4}{b^2} = 4225$ , and solving this equation for $b$ gives $b = \frac{a^2}{\sqrt{4225 - a^2}}$ .

If $a$ and $b$ are integers in the equation $b = \frac{a^2}{\sqrt{4225 - a^2}}$ for $a > 0$ and $b > 0$ , then $s = \sqrt{4225 - a^2}$ must be an integer. The only possibilities for $(a, s)$ with integer solutions for $a > 0$ are $(16, 63)$ , $(25, 60)$ , $(33, 56)$ , $(39, 52)$ , $(52, 39)$ , $(56, 33)$ , $(60, 25)$ , and $(63, 16)$ , and out of these possibilities the only resulting integer value for $b = \frac{a^2}{s}$ is $(60, 25)$ , which means $a = 60$ and $b = \frac{60^2}{25} = 144$ .

Therefore, the only possible side lengths of the rectangular paper are $60$ and $144$ , and its perimeter is $P = 2 \cdot 60 + 2 \cdot 144 = \boxed{408}$ .