The proposition expressed by the title is what I've found recently. Anyone who has or knows similar ideas please let me know. Other comments are also welcome.
A paraphrase of solipsism: no one is possible to know what anyone else knows. More precisely, for any $x$ and $y$, if $x$ is not $y$ then it is impossible that $x$ knows what $y$ knows. In formula: \[\forall x\forall y(x\neq y\rightarrow\forall p\neg\Diamond(K_{x}K_{y}p\lor K_{x}\neg K_{y}p))\]
Let's denote the formula by $S$. Assuming
(a) $\vdash\Box\varphi\rightarrow\varphi$, we have \[\forall x\forall y(x\neq y\rightarrow\forall p(\neg K_{x}K_{y}p\land\neg K_{x}\neg K_{y}p))\]
Let's denote it by $S'$. Thus $\vdash S\rightarrow S'$. Suppose $a\neq b$. From $S'$ it follows that
(1) $\forall p(\neg K_{a}K_{b}p\land\neg K_{a}\neg K_{b}p)$, and
(2) $\forall p(\neg K_{b}K_{a}p\land\neg K_{b}\neg K_{a}p)$.
From (1) we have
(3) $\neg K_{a}K_{b}p\land\neg K_{a}\neg K_{b}p$, which implies that
(4) $\neg K_{a}K_{b}p$
From (2) by substituting $K_{b}p$ for $p$ we have
(5) $\neg K_{b}K_{a}K_{b}p\land\neg K_{b}\neg K_{a}K_{b}p$, which implies that
(6) $\neg K_{b}\neg K_{a}K_{b}p$ Let's denote $\neg K_{a}K_{b}p$ by $P$. Then we have
(7) $\vdash S\rightarrow P$, and
(8) $\vdash S\rightarrow\neg K_{b}P$ Assuming the rule
(b) $\vdash \varphi\rightarrow\psi\Longrightarrow\ \vdash K\varphi\rightarrow K\psi$,
from (7) we have
(9) $\vdash K_{b}S\rightarrow K_{b}P$. Assuming
(c) $\vdash K\varphi\rightarrow\varphi$, it follows from (8) that
(10) $\vdash K_{b}S\rightarrow\neg K_{b}P$ Now (9) and (10) together imply that
(11) $\vdash\neg K_{b}S$. Assuming the rule
(d) $\vdash\varphi\Longrightarrow\ \vdash\Box\varphi$, we have
(12) $\vdash\neg\Diamond K_{b}S$ Note that $b$ can be arbitrary. Therefore,
(13) $\vdash\forall x\neg\Diamond K_{x}S$, which means that $S$ is unknowable. In arriving at the conclusion, we only assumed (a)--(d), which are all plausible for modal logic and epistemic logic.