Surjectivity of $a^x+x\pmod n$.

Question

Given positive integers $a,b$ and $n$, prove that there exists a positive integer $x$ such that $$a^x + x \equiv b \pmod n.$$ So I found this problem on the aops. You can find a couple of answers there, but one I personally prefer is the following, (I'll add in more details to it)

Let's proceed by strong induction, suppose $\forall k<n$ $\exists x$ such that $k\mid a^x+x-b$. Then since $\phi(n)<n$ there must be an $m$ such that $\phi(n)|a^m+m-b$. Write $a^m+m-b=T\phi(n)$ then choose $M=tn\phi(n)-(T\phi(n)-m)>0$ So $$a^M+M\equiv a^{tn\phi(n)-(T\phi(n)-m)}+tn\phi(n)-(T\phi(n)-m)\equiv a^m-m-T\phi(n)\equiv b\pmod n$$ This is all well and good till you realize that we used Euler's theorem and we need $(a,n)=1$. Unfortunately I was not able to fix this when $(a,n)>1$ But I think I have a solution when $a$ is a prime,

First I'll prove an important result when $a=p$ : If $x$ exists then there are infinitely many such $x$, if $p\nmid n$ then let $x_i=x+in\phi(n)$ and check that they all work. If $p\mid n$ write $n=p^rs$ with $(p,s)=1$ so $$p^{x_i}\equiv p^{x+in\phi(n)}\equiv p^x\pmod{p^rs}\iff p^{in\phi(n)}\equiv 1 \mod \frac{p^rs}{(p^r,p^x)}$$

I claim $r\le x$, indeed $r>x\implies p^x\mid p^rs=n$ but $p^x+x\equiv b\pmod n\equiv b\pmod{p^x}$ so $p^x\mid x-b$ absurd unless $x=b$ but that implies $n\mid p^x$ which is absurd. So $\iff p^{in\phi(n)}\equiv 1 \pmod s$ Which is true. Now we're ready to finish it off,

Let $a=p$, if $p\nmid n$ proceed as the first proof. If $p\mid n$ then write $n=p^rs$ with $(p,s)=1$. Therefore $$p^M=a^{\phi(n)(tn-T)+m)}\equiv p^m\pmod{p^rs}\iff p^{\phi(n)(tn-T)}\equiv 1\mod \frac{p^rs}{(p^r,p^m)}$$ Taking $m$ to be big enough (We're allowed to do that because of the above result) we get $\iff p^{\phi(n)(tn-T)}\equiv 1\pmod s$ which is true because $s\mid n\implies \phi(s)\mid \phi(n)$ so we can write $\phi(n)=\phi(s)q$ to get $$p^{q\phi(s)(tn-T)}\equiv 1\pmod s $$

Answer 1

You are trying to prove the claim using Euler's theorem and strong induction.

I think the following idea should help. (Yimin Ge is using the idea.)

The idea is to write $n$ as $de$ where $d,e$ are positive integers satisfying $\gcd(d,e)=1$ and $\gcd(d,a)=1$ where every prime divisor $p$ of $e$ divides $a$.

Proof :

We can write $n=de$ where $d,e$ are positive integers satisfying $\gcd(d,e)=1$ and $\gcd(d,a)=1$ where every prime divisor $p$ of $e$ divides $a$.

(Example : If $n=2^35^311^1$ and $a=2^23^25^17^1$, then $e=2^35^3$ and $d=11$.)

For $x$ large enough, we have $e\mid a^{x}$, and $x$ has to be of the form $x=b+ye$ where $y$ is a positive integer. Every $x$ large enough which has this form satisfies $a^{x}+x\equiv b \pmod e$.

If $y$ is a solution of $a^{b+ye}+b+ye\equiv b \pmod d$, then so is $y+\lambda d \varphi(d)$ for every positive integer $\lambda$. So, we can find a large enough $y$ such that $a^{b+ye}+b+ye\equiv b\pmod d$.

So, finally let us prove by strong induction on $d$ that there is a solution $y$ of $a^{b+ye}+b+ye\equiv b\pmod d$.

Suppose $\forall k<d$ $\exists y$ such that $k\mid a^{b+ye}+b+ye-b$. Then since $\phi(d)<d$ there must be an $m$ such that $\phi(d)|a^{b+me}+b+me-b$. Write $a^{b+me}+b+me-b=T\phi(d)$. Since $\gcd(d,e)=1$, there are integers $g,h$ such that $gd+he=1$. Then choose $M=td\phi(d)-T\phi(d)h+m\gt 0$. Using $a^{\phi(d)}\equiv 1\pmod d$, we get $$\begin{align}a^{b+Me}+b+Me&=a^{b+td\phi(d)e-T\phi(d)he+me}+b+td\phi(d)e-T\phi(d)he+me \\\\&\equiv b+a^{b+me}+me-T\phi(d)(1-dg)\pmod d \\\\&\equiv b+a^{b+me}+me-T\phi(d)\pmod d \\\\&\equiv b\pmod d.\ \blacksquare\end{align}$$

---

Added :

We define $d,e$ as $$e=\begin{cases}1&\text{if $\gcd(n,a)=1$}\\\\\displaystyle\prod_{p\ \in A}p^{\operatorname{ord}_{p}(n)}&\text{if $\gcd(n,a)\gt 1$}\end{cases}$$ and $d=\dfrac ne$ where $$A=\bigg\{p\ \bigg|\ \text{$p$ is a prime number which divides $\gcd(n,a)$}\bigg\}$$ and $\operatorname{ord}_{p}(n)$ is the largest exponent $a$ of the power $p^a$ of prime $p$ that divides $n$.

Then, such a pair $(d,e)$ always exists and satisfies the following three conditions :

• $\gcd(d,e)=1$

• $\gcd(d,a)=1$

• Every prime divisor $p$ of $e$ divides $a$.

Example :

If $n=2^35^311^1$ and $a=2^23^25^17^1$, then since $\gcd(n,a)=2^2\cdot 5^1\gt 1$, we have $A=\{2,5\}$.

Therefore, we get $$e=\prod_{p\ \in \{2,5\}}p^{\operatorname{ord}_{p}(n)}=2^3\cdot 5^3\qquad\text{and}\qquad d=\frac{n}{e}=11$$