We denote the Euler's totient as $\varphi(x)$, the Dedekind psi function as $\psi(x)$ and the sum of divisors function as $\sigma(x)$. Are well-known arithmetic functions, see Wikipedia. I would like to get idea about the veracity of the following conjecture (I have some computational evidence of it).
Conjecture. If $1\leq y<x$ are integers such that $2x=27y$, $$\psi(x+y)-\psi(x-y)=0,\tag{1}$$ $$\varphi(x+y)-\varphi(x-y)=2^8\tag{2}$$ and $$\psi(\sigma(x+y))-\psi(\sigma(x-y))=2^7\cdot 3^2\tag{3}$$ hold, then $x=1836$ and $y=136$.
As remark $1152/2=576$ divides $17^4-1$ (Wolfram Alpha online calculator), and $1152=\sigma(x)-\psi(x).$ Also $\psi(y)$ divides $\psi(x)$ and for the product of distinc primes dividing $x$ or $y$ (see Wikipedia Radical of an integer) we've $\operatorname{rad}(x)=2\operatorname{rad}(y)$.
Question. I would like to finish the reasoning to get a proof, or refute previous conjecture. Many thanks.
It's obvious that $3\mid x$. From the equation $(1)$ by cases for $y=2^{\alpha}\cdot 29^\beta\cdot5^\delta\cdot m$ with $m\geq 1$ integer such that $\gcd(2,m)=\gcd(5,m)=\gcd(29,m)=1$, for integers $\alpha\geq 1$, $\beta,\delta\geq 0$ then $\alpha\geq 1$ and $\beta=\delta=0$. By using second equation $(2)$ and the theory related to Fermat prime, powers of two and a theorem due to Gauss I know that I need to analyze the solutions of $\varphi(m)=2^{\lambda}$ (I got two cases $\varphi(m)=2^{6-\alpha}$ and $\varphi(m)=2^{7-\alpha}$), and from here my candidates are $y=2^{\alpha}\cdot 3^a\cdot 17^b$ with $a,b\in{0,1}$. If there are not mistakes my candidates for solutions after I've used $(1)$ and $(2)$ and $2x=27y$ are for $\alpha=3$ with $a=0$ and $b=1$, and secondly for $\alpha=6$ is $y=2^6\cdot 3$ that must be discarded.
As aside comment for the curious reader, one can to deduce dozens and dozends of similars equations for these integers $x=1836$ and $y=136$ involving previous arithmetic functions, and their iterates, quotients or compositions of functions. I've computational evidence for similar conjectures, for example the following conjecture.
Conjecture. If $1\leq y<x$ are integers such that $2x=27y$, $17\mid x$, $17\mid y$ and satisfy $(1)$, $(2)$ and $$\varphi\left(\frac{x+y}{17}\right)-\varphi\left(\frac{x-y}{17}\right)=\frac{x+y}{17}-\frac{x-y}{17}$$ then $x=1836$ and $y=136$.
Given that $2x=27y=3^3y$, we have $x=3^3z$ and $y=2z$ for some positive integer $z$. Plugging this in yields \begin{eqnarray*} \psi(29z)-\psi(25z)&=&0,\tag{1}\\ \varphi(29z)-\varphi(25z)&=&2^8,\tag{2}\\ \psi(\sigma(29z))-\psi(\sigma(25z))&=&2^7\times3^2\tag{3}. \end{eqnarray*} The first identity holds precisely for all positive integers $z$ that are coprime to both $29$ and $25$. Then the second identity reduces to $\varphi(z)=2^5$. This strongly bounds $z$ from above, and in fact this forces $$z\in\{51,64,68,80,96,102,120\},$$ We already found that $z$ is coprime to $5$, and then a routine check of the remaining $5$ numbers shows that only $z=68$ satisfies $(3)$.
As for your second conjecture; in this case $x=3^3\times17\times z$ and $y=2\times17\times z$ for some positive integer $z$. Again $(1)$ implies that $z$ is coprime to $25$ and $29$, and so $(2)$ implies that $\varphi(17z)=2^5$. Checking the list above we see that $z\in\{3,4,6\}$, of which only $z=4$ satisfies the last identity.