# Equation of the Chebyshev $\psi$ function

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?

• Taking exponentials, we get $l(n!)=l(A)l(A+2)$, where $l(x)$ is the least common multiple of all natural numbers below $x$. We have $A\approx n!/2$, so there will be a prime $p$ between $A+2$ and $n!$ by Bertrand for large $n$. But then $p\mid l(n!)$ and $p\nmid l(A)l(A+2)$. If anyone feels like writing up the exact bounds, feel free to turn this into an answer. Commented Apr 7, 2020 at 11:40
• Exact bound of what can you explain more Commented Apr 7, 2020 at 11:50
• Exact bound on how large $n$ has to be so that $n! > 2(A+2)$, so that Bertrans can be applied. Commented Apr 7, 2020 at 11:57
• So if we find this bound that mean when take a n and A greater than this bound the equation E does not has integers solution Commented Apr 7, 2020 at 12:05

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 $$\frac{m}{2} unless $$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 $$2^k$$.
• Very nice! I was thinking of a size argument, using some form of PNT, but your argument using the $2$-adic valuation is definitely nicer (and more elementary) Commented Apr 8, 2020 at 16:53
• @Wojowu Thanks! The key idea was yours. Commented Apr 8, 2020 at 17:08
• If p between $m$ and $2m$ how do you garante that p divide $m$ Commented Apr 8, 2020 at 17:20
• @Abdallahchaibeddrraa I am talking about the divisibility of $l(m)$ by $p$; recall that it is the l.c.m of integers $1,\dots,m$. Commented Apr 8, 2020 at 17:28