Consider $\Psi(x)$ to be the Chebyshev function given by 


$$\Psi(x)=\sum\_{n\leq x} \Lambda(n)$$


where $\Lambda(n)$ is the Mangoldt function which is equal 0 unless $n $ is prime power, and let $(E)$ be the following equation:


$$\Psi(n!)=\Psi(A) + \Psi(A+2) \tag E \, ,$$


where $n , A $ are integers and $A$ is even.


Can $(E)$ have integer solution or not? Can we relate it with other conjectures or open problems in number theory?



Equation of the Chebyshev $\psi$ function
This solution builds on @Wojowu's comment. The Mangoldt function $\Lambda(n)$ is defined as $\log p$ if $n=p^k$ is a prime power and as zero otherwise. The Chebyshev function
$$
\Psi(x)=\sum\_{n\leq x}\Lambda(n)
$$
is thus the logarithm of the least common multiple of $1,2,\dots,\lfloor x\rfloor$ because it could be written as $\sum\_{p\leq x\text{ prime}}\lfloor\log\_{p}x\rfloor\log p$. Its exponential $l:={\rm{e}}^{\Psi}$ is given by 
$l(x)=\log\big({\rm{lcm}}(1,\dots,\lfloor x\rfloor)\big)$.


I consider the more general equation 
$$\Psi(m)=\Psi(A) + \Psi(A+2) \tag {E1},$$
where $A$ and $m$ are positive integers. Exponentiating, one needs to solve $$l(m)=l(A)\,l(A+2)\tag {E2}.$$
If $A+2\leq\frac{m}{2}$, there is no solution: By Bertrand's Postulate (a.k.a Chebyshev's Theorem) there is a prime $p$ with $\frac{m}{2}<p<m$ unless $m\leq 2$. Such a prime divides the LHS of (E2) but neither of $l(A)$ and $l(A+2)$ on the RHS since $p>A+2$. It is easy to directly check that there is no solution with $m=1$.


So suppose $A+2>\frac{m}{2}$. The exponent of $2$ in the prime factorization of $l(m)$ is $\lfloor\log\_2 m\rfloor$ while the same numbers for $l(A)$ and $l(A+2)$ are given by $\lfloor\log\_2 A\rfloor$ and $\lfloor\log\_2 (A+2)\rfloor$ respectively. Therefore, (E2) implies
$$\lfloor\log\_2 m\rfloor=\lfloor\log\_2 A\rfloor+\lfloor\log\_2 (A+2)\rfloor.$$
But, in view of $A+2>\frac{m}{2}$:
$$
\lfloor\log\_2 (A+2)\rfloor\geq \left\lfloor\log\_2 \frac{m}{2}\right\rfloor=
\lfloor\log\_2 m\rfloor-1;
$$
which requires $\lfloor\log\_2 A\rfloor$ to be not greater than $1$. Hence the only choices for $A$ are $1,2,3$. Directly checking them, we observe that the only solutions to (E1) is $A=1,m=3$. 


**Added:** Changing $A+2$ to $B$ in (E1), the same idea could be used to study the equation 
$$
\Psi(m)=\Psi(A) + \Psi(B)
$$
where $A\leq B$. Again, $B>\frac{m}{2}$ (unless $A=B=m=1$) which implies
$\lfloor\log\_2 A\rfloor\leq 1$. So $A\in\{1,2,3\}$.


1. If $A=1$, then $\Psi(A)=0$ and there are infinitely many solutions, e.g. by setting $m=B$.
2. If $A=2$, then $\Psi(A)=\log 2$. The equation $\Psi(m)=\Psi(A)+\log 2$ admits infinitely many solutions; for instance, $m=2^k, A=2^k-1$ yields a solution since $${\rm{lcm}}(1,\dots,2^k)=2\times{\rm{lcm}}(1,\dots,2^k-1).$$
3. There exist solutions when $A=3$. For instance: 
$$\Psi(4)=\log 12=\log 2+\log 6=\Psi(2)+\Psi(3);$$
$$\Psi(10)=\Psi(9)=\log 2520=\log 420+\log 6=\Psi(7)+\Psi(3).$$
In general, solutions $(A=3,B,m)$ could be characterized with $B>\frac{m}{2}$ and the property that there should be exactly two prime powers in $\{B+1,\dots,m\}$, one in the form of $3^j$ and the other in the form of $2^k$.


