If $n$ is even,$a^n+1=x^2+1,(x=a^{\frac{n}{2}})$ So 3 doesn't divide $a^n+1$ and we see 3 doesn't divide $n!$ So $n=2$. But what if $n$ is odd,

We can use well-known (actually I never proved it with $a+1$ but rather $a-1$ but in this case I bet result is still valid and proof doesn't change much [induction]) if $p^b || a+1$ then $p^c || n$ is equivalent to $p^{b+c}||a^n+1$. Using that $v_p (n!)= \sum_{i=1} ^{\infty} [\frac{n}{p^i} ]$ we can compare $v_p(a^n+1)$ and $v_p(n!)$ and quickly bring whole problem to checking manually some small $n$.

No you cant since there are $\infty$ many $n$ with that property Or how would you do it¿

I would fix $n$ and assume that there exist such an $a$, then use above (from which we get bound on $n$) - is there really infinite number of solutions ? (can you show some infinite family of solutions) Edit: I've seen my mistake - I thought $0 \leq a < n$

Exactly the primes work But I will not post a proof that they work since this spoils half of the problem.

Nima Ahmadi Pour wrote: Find all $n$ such that there exists a unique integer $a$ such that $0\leq a<n!$ with the following property: \[ n!|a^n+1 \] If $n!$ divides $a^n+1$ then $gcd(n,a) = 1$. But $a < n$. Hence $a = 0$ or $1$. Then $n!$ can be $1$ or $2$, so $n$ is $1$ or $2$.

Surely not, just try $n=3,5,7,...$ I think you missinterpreted some places of $!$'s.

Sorry it's $a<n!$ I 'll try to come up with a solution

Check that for $n=2,3$ there is unique $a$ satisfying problem. Now assume $n \geq 4$. - If $n$ is even then $4|n!$. Hence $4|a^n+1$ and we get a contradiction. - If $n$ is odd composite then there are at least two such $a$'s: Obviously for $a=n!-1$ we have $n!|a^n+1$. Lets prove that if $a=(n-1)!-1$ then $n!|a^n+1$. Note that \[a^n+1=((n-1)!-1)^n+1=((n-1)!)^2A+(n-1)!n-1+1=((n-1)!)^2A+n!\] Because $n$ is composite $n!|((n-1)!)^2$ and so $n!|a^n+1$. Hence there are at least two such $a$'s. - If $n$ is prime then there is unique such $a$: Let $n!=p!=r_1^{a_1}r_2^{a_2}....r_k^{a_k}$ be the prime factorization of $n!$ and let $(r,b)=(r_i,a_i)$ for some $i$ so that $r^b|a^p+1$. Then $r^b|a^{2p}-1$ and from Euler's theorem we also have $r^b|a^{r^{b-1}(r-1)}-1$. So $r^b|a^d-1$ where $d=\gcd(2p,r^{b-1}(r-1))|2$ . If $d=1$ then $a^p+1\equiv 2 \mod r^b$ which is obviously a contradiction. So $d=2$ and $r^b|a^2-1=(a-1)(a+1)$. If $r$ is odd then then $a\equiv 1 \mod r^b$ or $a\equiv -1 \mod r^b$. As above we must have $a \equiv -1 \mod r^b$. If $r=2$ then taking $a=2k+1$ gives $2^{b-2}|k(k+1)$. If $k\equiv 0 \mod 2^{b-2}$ then $a^p+1=(2k+1)^p+1\equiv 2 \mod 2^{b-1}$. Since $b \geq 3$ this is a contradiction. So $k\equiv -1 \mod 2^{b-2}$ and $a\equiv -1 \mod 2^{b-1}$. If $a\equiv 2^{b-1}-1 \mod 2^b$ then \[a^p+1 \equiv (2^{b-1}-1)^p+1 \equiv 2^{b-1} \mod 2^b\] This gives $a\equiv -1 \mod 2^b$. Thus we proved that $a\equiv -1 \mod r_i^{a_i}$ for all $i$. From the Chinese remainder theorem this system of equations has unique solution in $[1, r_1^{a_1}r_2^{a_2}....r_k^{a_k})=[1, p!)$. Hence when $n$ is prime there is unique such $a$.

I have the same as cefer for evens and primes, but my proof for odd composite is a little different.

Note that $ n = 1, 2, 3$ are solutions. Suppose that $ n \geq 4$ is a solution, and let $ a$ be the unique integer. $ a$ and $ n!$ are relatively prime, since $ (a, n!) \leq (a^n, n!) \leq (a^n, a^n + 1) = 1$. I. The equation $ x^n \equiv_{n!} 1$ does not have a solution in $ 2 \leq x \leq n!$. Proof. Suppose the contrary, and let $ 2 \leq x \leq n!$ be a solution to that equation. Then $ (ax)^n \equiv a^n x^n \equiv - 1 \pmod {n!},$ hence $ ax \equiv_{n!} a$ hence $ x \equiv_{n!} 1,$ contradiction. II. $ n$ is prime. Proof. Suppose the contrary, and let $ q$ be the least prime divisor of $ n,$ so that $ 2q \leq n.$ Then $ q \,|\, \tfrac{n!}{q},$ and $ x = \tfrac{n!}{q} + 1$ satisfies $ x^n - 1 = \tfrac{n!}{p} (1 + x + \cdots + x^{n - 1})$ and also $ 1 + x + \cdots + x^{n - 1} \equiv_p 1 + 1 + \cdots + 1 \equiv_p 0,$ hence $ n!$ divides $ x^n - 1,$ contradiction. III. All primes are solutions. Proof Let $ p$ be a prime. Then $ a = p! - 1$ is a solution to $ p! \,|\, a^p + 1.$ Suppose that $ b$ is another solution, then (because both are relatively prime to $ p!,$) $ b^{ - 1}a \bmod p!$ is a solution to $ x^p \equiv_{p!} 1,$ then $ (x, p!) = 1$ so if $ \delta$ is the order of $ x$ modulo $ p!$ it must be $ \delta \, |\, p$ and $ \delta \,|\, \phi(p!),$ and since $ (p, \phi(p!)) = 1,$ $ \delta = 1,$ so $ x = 1.$ Hence $ b^{ - 1} a = 1$ so $ a = b$ and we have that $ a$ is indeed unique.

ONE MUST NOT FORGET: Also note that if $n=1$, the only possible $a$ is $a=0$ since $0 \le 0 \textless 1$ and it works, so $n=1$ works. Zhero wrote: Note that $ n \neq 1$, since we cannot find an integer between 0 and 1!. That is wrong since $n=1$ is possible.

Step 1. $n=1$ is a solution Step 2. For even $n$ we have that if $a$ satisfies then so does $n!-a$ so we must have $a=\frac {n!}{2}$ and for $n\ge 4$ we get that $n|1$ absurd. Step 3. $n=2$ is indeed a solution Step 4 For odd composite $n=pq$ where $p$ is prime and $q\ge 3$ we have $p^h||n!-1$ with $a\ge 2$ then easily find with LTE that we can choose more then one $a$ modulo $p^h$ such that $p^h|a^p+1|a^n+1$ giving using chineese remainder theorem to possibilities for $a$ mod $n!$. Step 5 For prime $n=p$ we prove that for each $q^h||n!$ and $q^h|a^p+1$ we have $q^h|a+1$ and again finish with chineese remainder theorem. Step 6 Acknowledge your awesomeness

We shall check for primes and composites seperately. Composite Consider $n \ge 3$.Then for even $n$ we have $4 \mid n! \mid a^n+1$.But remainders of $a^n+1$ modulo $4$ are $1,2$.So there do not exist such $a$. Consider odd $n$.It is obvious that $n! \mid (n!-1)^n+1$ so one of the choices for $a$ is $n!-1$.Also note that if $p$ is a prime divisor of $n$ then $n!|(\frac{n!}{p}-1)^n+1$ since each term on the RHS is a multiple of $n!$.(Note that if $p|n$ then $p^2 \mid n!$.Thus there are atleast two choices for $a$. Prime Let $p \ge 3$ be a prime $\le n$ where $n$ is another prime.If $p \mid \frac{a^n+1}{a+1}$ then $p \mid -(a^n+1)=(-a)^n-1$ and $p \mid (-a)^{p-1}-1 \Rightarrow p \mid (-a)^{\text{gcd}(n,p-1)-1}=-(a+1)$.We then see that $\frac{a^n+1}{a+1} \equiv n\pmod{p}$ which shows that $p=n$.We now also see that $\text{gcd}(\frac{a^n+1}{a+1},(n-1)!)=1$.Thus $n! \mid a+1$ and so by the inequality bound we see that $a=n!-1$ is the only choice.

Apologize for the bump, but I don't think I'm just restating anything here... We start by proving a lemma which is basically the cornerstone of orders and is quite an elegant result. Consider $(\mathbb{Z}/n\mathbb{Z}, \times)$, the multiplicative subgroup modulo $n$, or all numbers relatively prime to $n$ modulo $n$, I claim the following lemma: Lemma: If $f(x) = x^m$ takes $(\mathbb{Z}/n\mathbb{Z}, \times)$ to $(\mathbb{Z}/n\mathbb{Z}, \times)$, it is bijective iff $(m, \phi{(n)}) = 1$ Proof: Observe that the only if direction is trivial. Now we prove the if direction. I claim that if $x^m \equiv 1 \pmod{n}$, then $x \equiv 1 \pmod{n}$. Since ${x}^{\phi (n)}\equiv 1 \pmod {n}$, by the Euclidean Algorithm we can attain our result. Observe that the rest of the problem is now easy, as we now show that the function is injective. Assume otherwise. Then $a^m \equiv b^m \pmod{n}$, and letting $k = a/b \pmod{n}$, $k^m \equiv 1\pmod{n}$, so $k = 1$ and $a = b$. The function is injective, and since the domain and range have equal size, the function is bijective. $\blacksquare$ Assume that $n$ was prime. Then since $(a^n, n!) = 1$, $(a, n!)=1$ so $a$ is in $(\mathbb{Z}/n!\mathbb{Z}, \times)$. Hence, by the above lemma, as $(\phi{(p!)}, p) = 1$ (which can be easily verified), there is a unique value of $a$ such that $n! | a^n + 1$. Assume that $n$ was even and not $2$. Then observe that $4 | 4! | n!$, but $a^{2k}+1$ can never be a multiple of $4$, so we are done. Assume that $n$ was composite. Observe that in this situation, per the above ideas, take the smallest prime factor of $n$ and call it $p$. Observe that $p^2 | n!$. Hence, $p | \phi{(n!)}$ and $(n, \phi{n!}) \neq 1$, so the function is not injective. Hence, take $a^n = b^n \pmod{n!}$, for $a, b \in (\mathbb{Z}/n!\mathbb{Z}, \times)$, and $a \not \equiv b \pmod{n!}$. So $(a/b)^n \equiv 1 \pmod{n!}$ and $-(a/b)^n \equiv -1 \pmod{n!}$, but $a/b \not \equiv 1 \mod{n!}$, and $-1 \pmod{n!}$ is also a solution, so there are more solutions than just one in this case. Hence the answer is all $n$ prime and $1$.

We claim that the following statement holds. Let $n$ be a positive integer and $0 \le a <n!$ be an integer such that $n! \mid a^n+1$. Let $f(n)$ be the number of such integers $a$. Then we have $f(n)=1$ if and only if $n=1$ or $n$ is a prime number. Note that even numbers don't work after $n=2$ since $3 \mid (a^{\frac{n}{2}})^2+1$ is simply not possible. Also note that $n=1$ does work. We now show that primes work and odd composite numbers don't. Primes Let $p$ be a prime number. Clearly $a=p!-1$ works. We prove that it is the only integer in this range. Indeed, let $p! \mid a^p+1$.Let $r$ be a prime factor of $p!$ and $v_r(p!)=t$. Let $d$ be the order of $a$ mod $r$. By the fact that $r \mid a^{2p}-1$ we have $d$ as one of $\{1,2,p,2p\}$ but we see that $1$ and $p$ are absurd values and if it is $2p$ then $2p \mid r-1$ which is false as $r \le p$. Thus, $d=2$. Therefore, $r \mid (a+1)$ and so by LTE we have $v_r(a^p+1)=v_r(a+1)+v_r(p) \ge v_r(p!)$ and since for $r \not= p$, $v_r(p)=0$ we get that $v_r(a+1) \ge v_r(p!)$ and for $r=p$ we automatically have $v_r(a+1) \ge v_r(p!)=1$. This proves that $p! \mid (a+1)$ establishing uniqueness of $a$. Odd Composites Let $n$ be an odd composite number. Let $r$ be a prime factor of $n$. We show that there is a number $a$ which works other than $n!-1$ which obviously does. Now, ofc we get by LTE; $v_r(a^n+1)=v_r(a+1)+v_r(n) \ge v_r(n!)$ and so we set $a \equiv -1 ( \bmod r^{v_r((n-1)!)})$ for all $r \mid n$. Otherwise, we set $a \equiv -1 (\bmod r^{v_r(n!)})$ for all $r$ prime divisors of $n!$ but not $n$. This clearly gives a working $a$ and infact by CRT, generates at least $n$ such $a$'s in the range $[0,n!)$ and we conclude that more than one such number exist. This proves that odd composites are failed by the proposition. The result follows, Q.E.D.

We claim that the answer is when $n$ is prime. The question is just asking for which $n$ is there a unique residue solution $a$ to $$a^n\equiv -1\pmod{n!}.$$By taking mod 4, this has no solutions when $n$ is even for $n\geq 4$ since $-1$ is not a quadratic residue mod 4, so note that $n=2$ works and assume $n$ odd from now on. When $n$ is prime, say $n=p$, consider any prime $q\leq p$. Let $k$ be the largest positive integer such that $q^k\mid p!.$ By Chinese Remainder Theorem, it suffices to show that $$a^p\equiv -1\pmod{q^k}$$has a unique solution for $a\pmod{q^k}$ for all primes $q\leq p.$ Let $r$ be a primitive root mod $q^k$, since all odd prime powers have a primitive root. Then, this becomes $$a^p\equiv r^{\frac{(q-1)q^{k-1}}{2}}\pmod{q^k}.$$Clearly, if $a$ is not relatively prime to $q$, there are no solutions in that case, so suppose that $a\equiv r^s\pmod{q^k}$. Then, this becomes $$r^{ps}\equiv r^{\frac{(q-1)q^{k-1}}{2}}\pmod{q^k}.$$Since $r$ is a primitive root, this is equivalent to $$ps\equiv \frac{(q-1)q^{k-1}}{2}\pmod{(q-1)q^{k-1}}.$$Since by definition $q\leq p$, $p$ is relatively prime to $(q-1)q^{k-1}$, hence there exists a unique residue $s$ that satisfies this, hence proven. Now, consider if $n$ is not prime. Write $n=pk$ for a prime $p$ and an integer $k\geq 2.$ Then, let $s$ be the largest positive integer such that $p^s$ divides $(pk)!$. In particular, note that $s\geq 2$ since the factor has at the very least a $p$ and a $2p$. Now, consider $$a^{pk}\equiv -1\pmod{p^s}.$$Again, letting $r$ be a primitive root mod $p^s$. this is $$a^{pk}\equiv r^{\frac{(p-1)p^{s-1}}{2}}\pmod{p^s}$$If $a=r^v$, this is $$r^{vpk}\equiv r^{\frac{(p-1)p^{s-1}}{2}}\pmod{p^s}$$so $$vpk\equiv \frac{(p-1)p^{s-1}}{2}\pmod{(p-1)p^{s-1}}.$$However, the left side, the right side, and the modulus are all divisible by $p$, which means that there cannot be a unique residue $v$ that works, contradiction, so we are done.

The answers are $n=1,n=2$ and $n$ prime. First settle trivial case of $n$ even. If $n$ is even, -1 must be a quadratic residue mod all primes $<n$. So then if $n\geq 3$ 2 is a quadratic residue mod 3 which is wrong. Now we have $n$ is odd. If $n$ is not prime and not 1, then it can be split into $pr$ where $p$ is prime and $r\geq 2$. Hence we have $2p\le n$, so $v_p(n!)\geq 2$. Let $k=v_p(n!)$. Now since such a solution is unique, and there is at least one solution($a=p!-1$), then if there is more than 1 solution to $a^n\equiv -1$ mod $p^k$ in the range $[0,p^k-1]$ then there are more than 2 solutions(by CRT). Let us choose a primitive root $g$ of $p^k$. Then $g^{\frac{p^k-p^{k-1}}{2}}\equiv -1 \mod{p}$, and we only need to show that $xn \equiv \frac{p^k-p^{k-1}}{2} \mod{p^k-p^{k-1}}$ has more than 1 solution, which is obvious. Now if $n$ is prime, it works. Firstly if there is only one solution for each $p$ for $xn \equiv \frac{p^k-p^{k-1}}{2} \mod{p^k-p^{k-1}}$ if $k=v_p(n)$ we are done. This is because by CRT there will only be one solution mod $n!$. So, this is obvious for primes $<n$ since $n$ is coprime to $p^k-p^{k-1}$(coprime to $p$ and $p-1$). Clearly for $p=n$, $n$ is coprime with $n-1$ so we are done, since $v_n(n!)=1$.

One can use Hensel to generate another solution.

The answer is all primes and $1$ which is trivial. For evens $>2$ we have $n! \equiv 0 \pmod 4,$ so $a^n \equiv 3 \pmod 4$ but is a square, impossible. We check $n=2$ works. If $n$ is odd and is a prime $p$ then the order of $a \pmod p$ must divide $2p$ but not $p$ so it must be either $2$ or $2p.$ But it must also divide $\varphi(p!)$ so it is not $2p.$ Thus $a^2 \equiv 1 \pmod p$ so $a^p \equiv a \pmod p$ so $a \equiv -1 \pmod p$ and $a=p!-1$ is the only solution which works. If $n$ is odd and composite, we take $a=(n-1)!-1$ and since $n \mid (n-1)!$ we have $n! \mid (n-1)!^2$ so when we expand all terms of $a^n$ using the binomial formula all terms must be divisible by $n!$ except for the last two, but the second to last term is $(n-1)! \cdot n = n!$ so it is also divisible by $n!$ so $a^n \equiv -1 \pmod{n!}$ and $n! \mid a^n+1.$ However we can also just take $a=n!-1$ so these don't work and we are done.

$\textcolor{blue}{\text{Claim } n \hspace{0.2cm}\text{even doesn´t work}}$ We would have for some $p\mid n!$ prime $p\mid a^n+1$ so by Fermat Christmas Theorem $p=2$ or $p\equiv 1\pmod 4$ $\textcolor{blue}{\text{Claim } n \hspace{0.2cm}\text{odd and composite doesn´t work}}$ As $n$ odd $n\mid (n!-1)^n-1$ so $a=n!-1$ works

$\textcolor{blue}{\text{Claim } p \hspace{0.2cm}\text{prime work}}$ Let $q\mid p! \hspace{0.1cm}$ prime, so $a^{2p}\equiv 1\pmod q$ and $a^{q-1}\equiv 1\pmod q$ so $\operatorname{Ord}_q(a)\mid 2\gcd\left(p,\frac{q-1}{2}\right)$ so $\operatorname{Ord}_q(a)=2$ thus $q\mid a+1$, then $$\nu_q\left(a^p+1\right)=\nu_q(a+1)+\nu_q(p)=\nu_q(a+1)$$So $p!\mid a+1$ then $a=p!-1$ is the unique solution here

The answer are the $\boxed{\text{primes}}$. If $n$ is even and greater than $2$, we have \[a^n \equiv -1 \pmod{n!} \iff a^n \equiv -1 \pmod{4},\] which is impossible. Manually check that $n=2$ is valid, so assume $n$ is odd henceforth. If $n$ is prime, we have \[a^n \equiv -1 \pmod{n!}\]\[\implies a^{2n} \equiv 1 \pmod{n!}.\] Hence, the order of $a$ modulo $n!$ is divisible by $2n$ but not $n$. This means it is either $2$ or $2n$. Consider a prime $p \mid n!$. We have \[a^{p-1} \equiv 1 \pmod{p},\] so the order of $a$ modulo $n!$ is $\gcd(p-1, 2n) = 2$. This means that \[a^n \equiv a \equiv -1 \pmod{p!},\] fixing $a=p!-1$. If $n$ is odd and composite, $a=n!-1$ still is valid, but we claim $(n-1)!-1$ will also be valid, violating the uniqueness of $a$. Note that \[\nu_{p}(((n-1)!-1)^n+1) = \nu_{p}((n-1)!) + \nu_{p}(n) = \nu_p(n!). \] We are done.

We claim the answer is $\boxed{\text{all primes}}$. For all primes $n=p$, we claim the desired unique integer is $p!-1$, which clearly works. To prove this, take an arbitrary prime $q<p$, and note we require \[a^p \equiv -1 \pmod q \implies \operatorname{ord}_qa \mid 2p, q-1 \implies \operatorname{ord}_qa = 2 \implies q \mid a+1.\]We conclude using LTE to get \[v_q(p!) \leq v_q(a^p+1) = v_q(a+1) \implies p! \mid a+1 \implies a=p!-1.\] For all evens other than 2, notice that if $m$ works, $n!-m$ also works (and clearly $\tfrac{n!}{2}$ always fails), giving us two distinct values. For all odds, we simply take $n!-1$ and $(n-1)!-1$. $\blacksquare$

Nice training problem, the answer are all prime numbers. First note that if $n$ is even, for a solution $a$, $n!-a$ is also a solution, thus $a=\frac{n!}{2}$, which only works for $n=2$. For odd $n$ the only solution should be $n!-1$, for composite $n$, however take $\frac{n!}{p}-1$, for some prime $p$ dividing $n$, and see that for example by $LTE$, $v_p(a^n+1)=v_p(a+1)+v_p(n) \ge v_p(n!)-1+1=v_p(n!)$, or just factor out and see that the second paranthesis is also divisible by $p$. Now for prime $n$, any prime dividing $a^n+1$ divides $a^{2n}-1$, thus it's order divides $2n$, but as it also divides $p-1$, for primes $p\le n$ we get that it divides $2$, meaning together with $p \mid a^n+1$, that $p \mid a+1$ for any prime $p\le n$, and using $LTE$ or factorization again it is easy to see that $\frac{a^n+1}{a+1}$ does not contribute with any prime power, thus we get $n! \mid a+1$, which is the desired conclusion.

We claim that the answer is all prime $n$, with $a=n!-1$ the unique value of $a$ in those cases. First, we prove that this works. For $n=2$, $a=1$ is the unique integer that works. For larger $n$, assume that $n!=p_1^{e_1}\cdot p_2^{e_2}\dots p_k^{e_k}\cdot n$, where $p_1$ through $p_k$ are the primes less than $n$. For any prime $p<n$, we will show that $p^k\mid a^n+1$ implies that $a\equiv -1 \pmod{p^k}$. The order of $a\pmod{{p_i}^{e_i}}$ must divide $\gcd((p-1)p^{k-1},2n)$, which equals $2$ since $n$ is a prime greater than $p$. The order can't be $1$, so it must be $2$, which then implies that $a\equiv -1 \pmod{{p_i}^{e_i}}$. Then for the prime $n$, since $a^n\equiv a\pmod{n}$ by FLT, for $n\mid a^n+1$ we require $a\equiv -1 \pmod{n}$. Then by CRT over all of the prime powers dividing $n!$, there is only one solution. Now we must prove that other numbers do not work. Even values of $n>2$ fail since $4\mid n!$ but $a^n+1$ can only be $1$ or $2\pmod{4}$. For odd composite $n$, consider a prime $p$ dividing $n$, and let $k$ be the maximum value such that $p^k\mid n!$. If we let $a\equiv -1\pmod{p^{k-1}}$, we can apply LTE to get that $\nu_p(a^n+1)=\nu_p(a+1)+\nu_p(n)\geq k-1 +\nu_p(n)$. Since $p\mid n$, $\nu_p(n)\geq 1$, so $\nu_p(a^n+1)\geq k$ and $p^k\mid a^n+1$. Thus there are $p$ different values $\pmod{p^k}$ that are $-1\pmod{p^{k-1}}$, each of which could be used when applying CRT over all of the prime powers. This tells us that the number of solutions will be a multiple of $p$, but this can clearly never be $1$. Thus prime $n$ are the only solutions, as desired.

The solution set is $\boxed{n = 1 \text{ and } n \text{ prime.}}$ Lemma 1: $n \in \{1, 2\}$ work trivially. Lemma 2: If $n$ works and is even, $n = 2$. Proof: Assume otherwise, letting $n = 2k$. Then $3 \mid n! \mid (a^k)^2 + 1$, impossible by Fermat's Christmas Theorem. Lemma 3: All odd composite numbers don't work. Proof: Same as anantmudgal09, so copied from there: Let $n$ be an odd composite number. Let $r$ be a prime factor of $n$. We show that there is a number $a$ which works other than $n!-1$ which obviously does. Now, ofc we get by LTE; $v_r(a^n+1)=v_r(a+1)+v_r(n) \ge v_r(n!)$ and so we set $a \equiv -1 ( \bmod r^{v_r((n-1)!)})$ for all $r \mid n$. Otherwise, we set $a \equiv -1 (\bmod r^{v_r(n!)})$ for all $r$ prime divisors of $n!$ but not $n$. This clearly gives a working $a$ and infact by CRT, generates at least $n$ such $a$'s in the range $[0,n!)$ and we conclude that more than one such number exist. This proves that odd composites are failed by the proposition. Lemma 4: All odd primes work. Proof: Attached below. Due to the above, the solution set is confirmed. $\square$

\begin{answer} Only primes and 1 work as the value of $n$ \end{answer} \begin{proof} \textbf{Prime $ n$ gives unique solution} let $p! = p_1^{\alpha_1} \dots p_k^{\alpha_k}$ so we need solution to $a^p \equiv -1 \pmod{p_i^{\alpha_i}}$ Clearly we have that $\gcd(p,p_i^{\alpha_i})=1$ then clearly $a^p$ is a bijection and we are done, as it gives out unique values for each $p_i^{\alpha_i}$ and then we apply CRT to finish. \textbf{Composite $n$ do not work} let $n! =p^{\nu_p(n!) }\cdot k$ where p is chosen such that $p \mid n \Rightarrow n=p\cdot t$ \begin{claim} if a solution to $a^n \equiv -1 \pmod{p^{\nu_p(n!)}}$ multiple solutions exist. \end{claim} \begin{proof} So we need $a^{pt} \equiv b^{pt} \equiv -1 \pmod{p^{\nu_p(n!) }}$ to have solutions in distinct $a,b$. clearly $a,b$ have to be co prime to $p_i$ and getting $a^p \equiv b^p \pmod{p^{\nu_p(n!) }}$ works as i can raise both to $t$ therefore we need \begin{center} $\left(\frac{a}{b} \right)^{p} \equiv 1 \pmod{p^{\nu_p(n!)}}$ \end{center} then clearly taking primitive roots we seen that this has multiple solutions. \end{proof} this clearly shows that composite $n$ do not work. \end{proof}

The answer is all prime $n.$ (And $1$ as well. bruhhhhh) First, we verify that the only even $n$ that works is $n = 2.$ For $n = 2,$ only $a = 1$ works, while for even $n > 2,$ if $a$ works, then so does $n! - a,$ and since $\frac{n!}{2}$ clearly doesn't work, we get an even number of solutions. Henceforth assume that $n$ is odd. Verification that odd primes $p$ work: We want $a^p \equiv -1 \pmod{p!},$ or $(-a)^p \equiv 1 \pmod{p!}.$ We want to show that there is exactly one value of $a$ which satisfies this, namely $a = n!-1.$ This is equivalent to showing that the order of any number modulo $p!$ cannot be $p.$ In fact, $p \nmid \phi(p!)$ because $$\phi(p!) = p! \cdot \prod_{q < p, q \text{ is prime}} \left(1 - \frac{1}{q}\right)$$has all of its prime factors less than $p.$ This means that if $(-a)^p \equiv 1 \pmod{p!},$ then $-a \equiv 1 \pmod{p!},$ giving exactly one solution, as required. Proof that odd composites $n$ don't work: Once again we would like to solve $(-a)^n \equiv 1 \pmod{n!}.$ For simplicity let $b = n!-a,$ so that we want $b^n \equiv 1 \pmod{n!}.$ Obviously $b = 1$ satisfies this. The key claim is that $b = (n-1)! + 1$ also satisfies this, which immediately shows that $a$ is not unique. To see why. write $$b^n - 1 = (b-1)(b^{n-1} + b^{n-2} + \dots + b + 1) = ((n-1)! + 1 - 1)(b^{n-2} + \dots + b + 1).$$The left factor becomes $(n-1)!,$ and $(n-1)! \equiv 0 \pmod{n}$ since $n$ is composite, so the right factor is $1^{n-2} + \dots + 1 + 1 \equiv 0 \pmod{n}.$ So the entire product is divisible by $n!,$ implying that $a = (n-1)! + 1$ works, as claimed. Thus all composite $n$ don't work. Therefore, only prime $n$ work, as desired. Remark: A simpler proof that even $n > 2$ don't work is that $n! \equiv 0 \pmod{3}$ while $a^n + 1 \equiv 1,2 \pmod{3}.$

It is easy to see that $\boxed{n=1}$ work. We consider the case for $n$ even, so $a^n+1\equiv 2\pmod{3}$ so $\boxed{n=2}$. If $n$ is odd we consider $2$ cases, for odd primes and odd composites, here $a=n!-1$ satisfies the conditions of the problem, now we check for $a=(n-1)!-1$, $((n-1)!-1)^n+1\equiv -1+1 \equiv 0\pmod{n!} $, here we used the fact that $((n-1)!)^r\equiv 0\pmod{n!}, r\geq 2$, but note that this contradicts the uniqueness of $a$. If $n$ is an odd prime $\text{ord}_{n!}(a)=2, 2n$, if $\text{ord}_{n!}(a)=2$ then $n!\mid a+1$ so a valid construction for $a$ can be $n!-1$. Hence $\boxed{\text{all odd primes n}}$ work. For $\text{ord}_{n!}(a)=2n$ we let $p$ be a prime such that $p\mid n!$, then $\text{ord}_{n}(a)\mid p-1$ which is simply not possible.

cool nt