The answers are $\boxed{(2,2,2)}$ and $\boxed{(3,4,3)}.$ First verify $1 < (1+p)^{1/p} < 2$ for any prime $p$ so $b = 1$ impossible. If $2\le b < p$ then take any prime $q\mid b!+p = a^p.$ Then $q > b.$ But $b! + p \le b^p < q^p \le a^p$ when $2 \le b < p,$ impossible. If $b \ge p$ then $a = kp$ for some positive integer $k.$ Note $v_p(a^p - p) = 1$ so $b < 2p.$ Note $k < p$ to allow $(kp)^p \le (2p-1)! + p.$ If some prime $q < p$ divides $k$ then $b < q,$ but $(kp)^p > (q-1)! + p,$ contradiction. So $k = 1.$ Note that $v_2(p^p-p) \le 2\log_2(p-1)$ (LTE) and $v_2(p!) \ge p-\log_2(p+1)$ (Legendre's). So $p-\log_2(p+1) \le 2\log_2(p-1) \implies p < 10.$ Finite case check on $p=2,3,5,7$ finishes. $\blacksquare$

Motivation: I have seen 2018 P5, 2019 P4. Think about size in addition to $v_p.$ Suppose $p>10.$ $\textit{(i)}$ If $b\geq 2p$ then $v_p(b!)\geq 2$ therefore $v_p(b!+p)=1$ hence $v_p(a^p)=1.$ This means that $p\mid a^p$ which implies that $p\mid a$ therefore $v_p(a^p)=pv_p(a)=p.$ Hence $p=1$ but this is absurd. $\textit{(ii)}$ If $b<p.$ In this case $b!<b^p\Rightarrow b!+b<b^p+p\Rightarrow a^p<b^p+p.$ Since perfect $p$ powers should be at least $p+1$ apart we have that $a^p<b^p\Rightarrow a<b.$ This implies that $a\mid b!$ therefore $a\mid p$ and either $a=1$ or $a=p.$ $a=1$ doesn't work. If $a=p$ then $p^p=b!+p$ so a size argument gives an immediate contradiction. $\textit{(iii)}$ If $p\leq b<2p.$ If there exists a prime $q$ such that $q\neq p$ and $q\mid a$ then $q>b\geq p.$ Furthermore, $a^p\geq q^pp^p>p^{2p}>(2p-1)!+p>b!+p.$ Hence no such $q$ can exist. Therefore $a=p^\alpha.$ By checking this size argument you see that we must have $\alpha=1.$ Thus $a=p.$ Using LTE we have $v_2(p^{p-1}-1)=v_2(p-1)+v_2(p+1)+v_2(p-1)-1=v_2(b!).$ Using that legendre formula or whatever it's called we get $v_2(b!)=b-s_2(b).$ Since we have $p\leq b<2p$ we get $v_2(b!)=b-s_2(b)\geq p-\log_2(2p).$ On the other hand $2\log_2(p-1)+\log(p+1)-1\geq 2v_2(p-1)+v_2(p+1)-1.$ Putting this stuff together we get $$2\log_2(p-1)+\log_2(p+1)-1\geq p-\log_2(2p)$$But this does not hold for $p>10.$ Thus we get our desired contradiction. Now we check $p=2,3,5,7$ with $b$ in the appropriate range.

Very easy for a 5. I found this comparable to P4. All answers are $(a,b,p) = (2,2,2), (3,4,3)$. The key claim is the following. Claim: $a$ is not divisible by any prime $q<p$. Proof: Assume so. Then, $q\nmid a^p-p=b!\implies b<q$. Therefore, $$b! \leq (q-1)! < q^{p-1} < q^p-p,$$so $a < q$, a contradiction. $\blacksquare$ Now we have two cases: If $p \nmid a$, then $p\nmid b!\implies b < p$, so $$b! \leq (p-1)! < p^{p-1} \leq p^p-p,$$or $a<p$. This gives a contradiction to the above claim. If $p\mid a$, then $p^2\nmid a^p-p \implies b < 2p$, so \begin{align*} b! &\leq (2p-1)! \\ &= (1(2p-1))(2(2p-2)) \dots ((p-1)(p+1)) p \\ &< (p^2)(p^2)\dots (p^2) p \\ &= p^{2p-1} \leq (p^2)^p-p, \\ \end{align*}so $a < p^2$. Therefore, the claim forces $a=p$. If $p\geq 5$, then by Zsigmondy theorem, we can find a prime $q$ such that $\operatorname{ord}_q(p) = p-1$, so that $q\geq 2(p-1)+1 = 2p-1$ and $q\mid p^p-p$. This forces $p^p-p=(2p-1)!$, and we can finish by bounding because the LHS is too small. In conclusion, we must have $a=p$ and $p\in\{2,3\}$.

squareman wrote: The answers are $\boxed{(2,2,2)}$ and $\boxed{(3,4,3)}.$ First verify $1 < (1+p)^{1/p} < 2$ for any prime $p$ so $b = 1$ impossible. If $2\le b < p$ then take any prime $q\mid b!+p = a^p.$ Then $q > b.$ But $b! + p \le b^p < q^p \le a^p$ when $2 \le b < p,$ impossible. If $b \ge p$ then $a = kp$ for some positive integer $k.$ Note $v_p(a^p - p) = 1$ so $b < 2p.$ Note $k < p$ to allow $(kp)^p \le (2p-1)! + p.$ If some prime $q < p$ divides $k$ then $b < q,$ but $(kp)^p > (q-1)! + p,$ contradiction. So $k = 1.$ Note that $v_2(p^p-p) \le 2\log_2(p-1)$ (LTE) and $v_2(p!) \ge p-\log_2(p+1)$ (Legendre's). So $p-\log_2(p+1) \le 2\log_2(p-1) \implies p < 10.$ Finite case check on $p=2,3,5,7$ finishes. $\blacksquare$ :D

The solutions are $(2,2,2)$ are $(3,4,3)$. 1. $p|a$: If not, then $b<p$, by $\mod p$, and $b<a$, by $\mod a$. Then, $a^p\geq a^{b+1}>2b!$ and $a^p\geq 2^p\geq 2p$, so $a^p$ is too large. 2. $p\leq b\leq 2p-1$: $b!$ must be a multiple of $p$, but not of $p^2$, which is the same as $p\leq b\leq 2p-1$. 3. $a=p$: Let $a=pk$. If $1<k<p$, then, by $\mod k$, $k|p$, contradiction. If $k\geq p$, then $(p^2)^p\leq a^p=b!+p\leq (2p-1)!+p\leq (2p-1)!\cdot p=\prod_{i=0}^{p-1} (p-i)(p+i)< (p^2)^p$, contradiction. Thus, $k=1$. 4. $p<5$: Suppose otherwise. Then, $p^p>p!+p$, so $b\geq p+1$. Now, our equation is $p(p^{p-1}-1)=b!$. But, by Lifting the Exponent Lemma, $v_2(p(p^{p-1}-1))=v_2(p-1)+v_2(p+1)+v_2(p-1)-1$. On the other hand, letting $p-1=d_1d_2$ (since $p\geq 5$ implies $p-1$ not prime), we have $v_2(b!)\geq v_2(d_1)+v_2(d_2)+v_2(p-1)+v_2(p+1)= v_2(p-1)+v_2(p-1)+v_2(p+1)> v_2(p(p^{p-1}-1))$, contradiction. Thus, $a=p=2$ or $a=p=3$, giving us the mentioned solutions.

This may actually be the single least interesting-looking IMO problem I've read so that's fun. $(2, 2, 2)$ and $(3, 4, 3)$. Discard $a = 1$. Notice that $\gcd(a, b!) \mid p$. Consider two cases depending on whether $p$ divides $a$. If $p \nmid a$ then $p \nmid b! + p$ so $b < p$. Now if $a \le b$ then $a$ and $b!$ share some prime factor which must be less than $p$, impossible. Therefore $a \ge b + 1$ and $a^p \ge (b + 1)^p > b^p + pb^{p - 1} > b! + p$ so we fail in this case. If $p \mid a$ then $p \mid b! + p$ so $b \ge p$. Moreover $p^2 \nmid a^p - p$ so $b < 2p$. Now let $a = pk$. Since $a$ cannot have prime factors in common with $b!$ other than $p$ we must either have $k = 1$ or $k \ge p$. In the latter case we have $a^p - p \ge p^{2p} - p > (2p - 1)! \ge b!$ which is a contradiction, so we must have $k = 1$ and $a = p$. We can also check $p^{2p} - p > p!$ so $b \ge p + 1$. Seeking a final contradiction we look at $\nu_2$. By LTE we have $\nu_2(p^p - p) = 2\nu_2(p - 1) + \nu_2(p + 1) - 1$. If $p \ge 7$ then from $b \ge p + 1$ we get $$\nu_2(b!) \ge \nu_2((p + 1)!) \ge \nu_2(p + 1) + \nu_2(p - 1) + \nu_2((p - 1)/2) + \nu_2(2) = 2\nu_2(p - 1) + \nu_2(p + 1)$$ which contradicts $b! \mid a^p - p$. We are then left to check $p = 2, p = 3$ and $p = 5$ which give the aforementioned solutions.

Really doable NT, and the ideas are very similar to those in 2021 N5 as well. Here is my solution: Case 1. Let $1<b \leq p-1$. Note that if a prime $q$ divides $a$, then $q \neq p$ and hence it's impossible $b \geq q$. Therefore, all prime divisors of $a$ are bigger than $b$, so $a>b$ and $b!+p \leq b^p < a^p$, done (the first inequality is due to $b<p$). Case 2. Let $p \leq b <2p$ (the other cases are impossible, since the RHS will have $v_p=1$). Obviously $p|a$, so let $a=cp$. The idea is that $c=1$ is the only possibility. Firstly, we can bound $c<p$ (we need this in order to get a prime $q \leq b$ divdiding $a$, so as to finish as in the above case), otherwise $(2p-1)!+p=a^p \geq p^{2p}$, but this is impossible, since $(2p-1)! \leq (\frac {2p-1}{2})^{2p-1} <p^{2p-1}$ by AM-GM, absurd (by the way, a similar bound with AM-GM was used in 2021 N5 to bound a factorial). Now take a prime $q|c$ and note that since $q<p<b$, $q$ will divide $b!$, hence $p$, absurd. So the only left possibility is $c=1$. We have the equation $p(p^{p-1}-1)=b!$. The idea is to use $v_2$ and Legendre (well, $v_2$ is easier to bound than $v_q$ for a prime $q|p-1$). Note that $v_2(b!)>\frac{b}{2}-1>\frac{p-1}{2}$ and $v_2(p^{p-1}-1)=2v_2(p-1)+v_2(p+1)-1<2log_2(p-1)$, and now finite case check shall suffice.

It was about time that an application of the Lifting the Exponent Lemma appears at IMO! (And indeed, as VicKmath7 pointed out, this feels too much at home if ISL 2021 N5 is fresh in your mind.) If $p=2$, then $b\geq 3$ is impossible by mod $3$ and only $a=b=2$ works. If $p=3$, then $b\geq 7$ is impossible by mod $7$ and only $a=3$, $b=4$ works. From now on, let $p\geq 5$. We shall repeatedly use the bound $n! \leq (\frac{n}{2})^n$ for $n\geq 6$ which follows by induction since $n+1 \leq \frac{(n+1)^{n+1}}{2n^n} \Leftrightarrow (1+\frac{1}{n})^n \geq 2$ holds by opening the brackets on the left and taking only the first two terms into account. Now suppose $p$ does not divide $a$. Clearly $a=1$ is impossible. Observe that $p$ does not divide $b!$ and so $b<p$, whence $a^p \leq (p-1)! + p < p!$ and so $a<p$. If $b=1$, then $1 = a^p - p \geq 2^p - p \geq 2$, contradiction, so $b\geq 2$. Hence if $a < b$, then $a$ would divide both $a^p$ and $b!$ and hence $p$, contradiction as $p$ is prime. So $a>b$, thus (since $p>b \geq 2$) $b! > b^p - p > b^b - b$ (as $b^b(b^{p-b}-1) > 2^{p-b} - 1 \geq p-b$) and $(b-1)! > b^{b-1} - 1$ which is false since $b^{b-1} \geq (b-1)!$ for $b\geq 2$ (both sides have the same number of multipliers, the left-hand one's all are greater). Hence there are no solutions when $p$ does not divide $a$. Next, consider $a = p^kx$ where $p$ does not divide $x$ and $k\geq 1$. Then $p^{kp}x^p = b! + p$. If $b\geq 2p$, then $b!$ and the left-hand side would be divisible by $p^2$ but $p$ would not be, so impossible. Hence $b<2p$ and so (using the above factorial bound) $b! + p \leq (2p-1)! + p \leq (\frac{2p-1}{2})^{2p-1} + p < p^{2p-1} + p < p^{2p}$. In particular, $p^{kp}x^p < p^{2p}$ and so $k\geq 2$ is impossible; also for $k=1$ we have $x<p$. Now if $x>1$, then either $x\leq b$ and hence $x$ divides $p^{p}x^{p} - b! = p$, impossible; or $x>b$ and so $p^pb^p < b! + p < p^{2p}$, i.e. $b<p$ -- but then in $p^{p}x^p = b! + p$ the term $b!$ would not be divisible by $p$ since $p$ is prime, contradiction. So it remains to deal with (the most interesting) $p^p = b! + p$ for $p\geq 5$. Again $p \leq b < 2p$. Write the equation as $p(p^{p-1} - 1) = b!$. The right-hand side is divisible by $2^{\frac{p+1}{2}}$ since $b\geq p-1$ and each of $2$, $4$, $\ldots$, $p-1$ is even while for $p\geq 5$ we have the multiplier $4$ which is divisible by $4$. On the other hand, by the Lifting the exponent lemma for the prime $2$, the power of $2$ in the left-hand side is $2\nu_2(p-1) + \nu_2(p+1) - 1$. If $p\equiv 1 \pmod 4$, then we obtain $2\nu_2(p-1) \geq \frac{p+1}{2}$, whence $4\log_2 p > p$ and $p^4 > 2^p$, which fails for $p\geq 17$ by a straightforward induction. If $p\equiv 3 \pmod 4$, then we obtain $\nu_2(p+1) + 1 \geq \frac{p+1}{2}$, whence $2\log_2(p+1) \geq p-1$ and $(p+1)^2 \geq 2^{p-1}$ which fails by induction for $p\geq 11$. Now we only have to check $p=5, 7, 13$. Well, $5^5 - 5 = 3120$ is divisible by $13$ but not by $9$, also $7^7 - 7$ is divisible by $7$ but not by $5$ and $13^{13} - 13$ is divisible by $61$ but not by $19$.

The answer is $\boxed{(a,b,p)=(2,2,2), (3,4,3)}$. These work. It remains to show they are the only solutions. If $a=1$, then $b!+p=1$, absurd. So assume $a>1$. Claim: $b<2p$. Proof: Suppose not. Then $p^2\mid b!$, which implies $\nu_p(b!+p)=1,$so it can't be a $p$th power. $\square$ Case 1: $a=p$. Then\[b! = p^p-p = p(p^{p-1} - 1)\]If $p=2$, then $b=2$, giving $(2,2,2)$ If $p=3$, then $b=4$, giving $(3,4,3)$. If $p=5$, then $b!=3120$, not possible. If $p=7$, then $b! = 823536$, not possible. Now assume $p>10$. By Zsigmondy's, we can find a prime dividing $p^{p-1}-1$ that doesn't divide $p^k -1$ for any $k<p-1$. So in other words, there exists a prime $q$, such that the order of $p$ mod $q$ is $p-1$. Claim: $q>2p-2$. Proof: Assume not. Since $p\ne q$, we have $p^{q-1}\equiv 1\pmod q$, so $p-1\mid q-1$. By our assumption that $q\le 2p-2$, we have $q=p-1$, which is not prime. $\square$ Now, we have $q\mid p^{p-1}-1\mid p^p - p = b!$. Since $q$ is prime, $b\ge q\ge 2p-1$. Because $b<2p$, this implies $b = 2p-1$. So $p^p - p = (2p-1)!$. Claim: $(2n-1)! > n^n$ for any positive integer $n>2$. Proof: We have\[(2n-1)! = (n-1) ! \cdot n(n+1)(n+2)\cdots (2n-1)\]Notice how $n(n+1)(n+2)\cdots (2n-1)$ is the product of $n$ terms, each of which are at least $n$. So $n(n+1)(n+2)\cdots (2n-1)\ge n^n$, which implies $(2n-1)! \ge (n-1)! n^n > n^n$, as desired. $\square$ Thus, $(2p-1)! > p^p > p^p - p$, contradiction. Case 2: $b<p$. Then we have $p\nmid b!$, so $p\nmid a$. Subcase 2.1: $a\le b$. Then $a\mid b!$, so $a\mid a^p - b! = p$, which implies $a=p$ (since $a>1$), absurd. Subcase 2.2: $a>b$. If $p=2$, then $b=1$, so $a^2 = 3$, not possible. Now assume $p>2$. Claim: For any positive integer $n>2$, we have $2^n > 2n$. Proof: We do this by induction on $n$ (base cases $3,4$). If $2^{n-1}> 2(n-1)$, then\[2^n = 2(2^{n-1}) > 4(n-1)>2n,\]as desired. $\square$ So $a^p \ge 2^p >2p$, which gives $b! > p$. Claim: For any positive integer $n>2$, we have $n^n > 2\cdot n!$. Proof: $2\cdot n!$ can be rewritten as $2\cdot 2\cdot 3\cdot 4\cdots n$. This is a product of $n$ terms, all of which are at most $n$, and at least one term is less than $n$. So $2\cdot n! = 2\cdot 2\cdot 3\cdot 4\cdots n<n^n$. $\square$ So\[a^p > b^p > b^b > 2\cdot b! > b! + p,\]contradiction. Case 3: $b\ge p$ and $a\ne p$. Since $p\mid b!$, we get $p\mid a$. Let $a=kp$ with $k>1$. Then\[(kp)^p = b! + p\] Claim: For any positive integer $n>1$, we have $n^{2n} > (2n-1)! + n$. Proof: First note that if $x$ and $y$ are positive, $x,y\ne n$, and $x+y = 2n$, then $xy<n^2$ by AM-GM. So\begin{align*} (2n-1)! = (1\cdot (2n-1)) \cdot (2\cdot (2n-2)) \cdots ((n-1)\cdot (n+1)) \cdot n\\ < (n^2)^{n-1}\cdot n \\ = n^{2n-1} \\ \le n^{2n}-n \\ \end{align*}$\square$ If $k\ge p$, then $(kp)^p \ge p^{2p} > (2p-1)! + p \ge b! + p$, contradiction. So $k<p$. Let $q$ be a prime divisor of $k$. Since $q<p\le b$, we have $q\mid b!$. So $q\mid (kp)^p - b! = p$, which implies $q=p$, contradiction. We have exhausted all cases, so the solutions described in the beginning are the only ones.

For $p=2,3$ then $(a,b,p)=(2,2,2)$ or $(3,4,3)$. So, assume $p>3$. Rewite as $a^p-p=b!$ We claim $b \geq p$. Assume $b<p$. If we let $p_1,p_2,\ldots,p_n$ be the primes $\leq b$ then $p_i$ divides $b!=a^p-p$ for every $1 \leq i \leq n$. If $p_i$ divides $a$ then $p_i$ divides $(a)^p-(a^p-p)=p$ which is clearly contradiction. Thus, $a \geq p_{n+1}$. So, \[ b! = a^p-p \geq p_{n+1}^p-p \geq (p_{n+1})! \]Hence, $b \geq p_{n+1}$ which is contradiction. Now notice that $p$ divides $b!=a^p-p$. Thus, $p$ divides $a$. We claim $b < 2p$. Indeed, notice that \[ 1=v_p(a^p-p)=v_p(b!) \]So, $b < 2p$ as desired. Now, we claim $p$ is the smallest prime divisor of $a$. Indeed, assume some $q<p$ divides $a$. $q < p \leq b$ so $q$ divides $b!=a^p-p$. Thus, $q$ divides $p$ which is contradtion. Hence, if $a>p$ then it must be the case that $a \geq p^2$. So, \[ p^{2p}-p \leq a^p-p = b! < (2p)!\]Which cannot hold. So, $a=p$. plugging back we get \[ p^p-p=b! \]Now, by LTE we have \[ v_2(b!)=v_2(p^p-p) = 2v_2(p-1)+v_2(p+1)-1\]Notice that $p^p-p>p!$ so $b \geq p+1$. Thus, \[ v_2(b!) \geq v_2((p+1)!) > v_2(p+1)+v_2(p-1)+ v_2 \left ( \frac{p-1}{2} \right ) +1\]which is strictly larger than \[ 2v_2(p-1)+v_2(p+1)-1 \]which gives contradiction. $\blacksquare$

The answer is $(2,2,2)$ and $(3,4,3)$ which clearly work. Clearly $b=1$ won't work for size reasons, so assume $b>1$. From AM-GM, for all $n$ we have $n!<(\tfrac{n+1}{2})^n<n^n$. First suppose that $b<p$, so $p \nmid b!$ and thus $p \nmid a$. By our bound, we require $a<b$, else $a^p>a^b\geq b^b>b!$, but then $a \mid b!-a^p$ and not $p$: contradiction. Thus suppose that $b \geq p$. Since $\nu_p(a^p-p)=1$, we must also have $b<2p$. Then we require $a<p^2$, else $a^p\geq (p^2)^p=p^{2p}\geq (\tfrac{b+1}{2})^{2p}>(\tfrac{b+1}{2})^b>b!$. Letting $a=kp$ where $k<p$, it follows that $k \mid a^p-b!$ but not $p$ unless $k=1$, so we reduce to the case where $a=p$. In this case, we count $\nu_2$: $\nu_2(p^p-p)=\nu_2(p^{p-1}-1)=\nu_2(p^2-1)+\nu_2(\tfrac{p-1}{2})=\nu_2(\tfrac{(p-1)^2(p+1)}{2})\leq \log_2(\tfrac{(p-1)^2(p+1)}{2})$ by exponent lifting, and $\nu_2(b!)\geq b-\log_2(b+1)\geq p-\log_2(p+1)$ by Legendre's. Thus we require $$p-\log_2(p+1)\leq \log_2(\tfrac{(p-1)^2(p+1)}{2}) \implies 2^p \leq \frac{(p-1)^2(p+1)^2}{2}.$$This fails for $p=15$ (yes this isn't a prime, no it doesn't matter), and by derivatives it also fails for all $p>15$, so we reduce to a finite case check. From this it's easy to get that the solutions provided are the only ones—instead of computing $p^p-p$, find a divisor of it that can't divide $b$ for each $p$. $\blacksquare$ The actual expressions are probably wrong but it doesn't really matter

Answer. $(a,b,p)=(2,2,2)$ and $(a,b,p)=(3,4,3)$. Those can be easily verified as solutions. Since $a^p=b!+p>1$ then $a>1$. Moreover $p\ge 2$ so $a^p\ge 2^p\ge p+2$. Hence $b\ge 2$. If $2\le b<p$, pick any prime divisor $q$ of $b!+p$. Hence $q>b$ and $q\mid a$. It can be proved by induction that $x^y\ge x+y$ for integers $x,y\ge 2$. Moreover for integers $x,y\ge 2$ we have $xy=(x-1)(y-1)-1+x+y\ge x+y$ so $$b^p=b^{b-1}b^{p-b+1}>b!p\ge b!+p=a^p\ge q^p,$$contradiction. If $b\ge p$ then $p\mid b!+p=a^p$ so $p\mid a$. Note that $v_p(a^p-p)=1$ so $b<2p$. Hence \begin{align*} a^p&\le (2p-1)!+p\\ &<(1)(2p-1)(2)(2p-2)\cdots(p-1)(p+1)p+p\\ &\le p[(p^2-1)^{p-1}+1]\\ &\le p^{2p-1}. \end{align*}Hence $a<p^2$. Moreover, if there exists a prime $q<p$ which divides $a$ then $\gcd(q,b!+p)=1$ because $b\ge p$. But $q\mid b!+p$, contradiction. Hence $a=p$ so $p^p-p=b!$. If $p=2$ we see that $b=2$ so we get the solution $\boxed{(a,b,p)=(2,2,2)}$. So now assume $p\ge 3$. By LTE and the fact that $\gcd(p-1,p+1)=2$ we obtain \begin{align*} v_2(p^p-p)&=v_2(p^2-1)+v_2\left(\frac{p-1}{2}\right)\\ &=\max\left(2v_2(p-1),v_2(p+1)+1\right)\\ &\le\max\left(2\log_2(p-1),\log_2(p+1)+1\right)\\ &=2\log_2(p-1). \end{align*}Moreover $$v_2(b!)\ge v_2(p!)=p-s_2(p)\ge p-\log_2(p+1).$$Hence $2\log_2(p-1)\ge p-\log_2(p+1)$. Equivalently $(p-1)(p^2-1)\ge 2^p$. Therefore $2^p<p^3$ so $p<10$. We are left to check $p=3,5,7$. If $p=3$ we obtain a solution $\boxed{(a,b,p)=(3,4,3)}$. If $p=5$ then $p^p-p=5^5-5=3120$ and $6!<3120<7!$. Hence no solution in this case. If $p=7$ then $$p^p-p=7^7-7\equiv 2^7-2=126\equiv 1\pmod{5}$$Hence $5\nmid 7^7-7$ so for a solution we need $7^7-7\le 4!$, but $7^7-7>7^2>4!$. Therefore there is no solution also in this case.

mathisreaI wrote: Find all triples $(a,b,p)$ of positives integers with $p$ prime and $$a^p=b!+p$$ Solution- Claim 1 $b\geq p$ If possible let $b<p$. If $a\leq b$, then $a|b! \implies a|p$, but $a\leq b<p$ gives $a=1$ which is not possible. Hence $a>b$, so $$a^p - p = (a-1+1)^p-p > (a-1)^p \geq b^p > b!$$Which is a contradiction, hence $b\geq p$ Claim 2 $a = p$ Going mod $p-1$, we have $a^p \equiv 1$, hence order of $a$ is $1$ or $p$. But order must divide $\phi(p-1)<p$ and hence it is $1$. Moreover $b\geq p \implies p|b! \implies p|a \implies a\equiv p \text{ mod} (p^2-p)$. Hence if $a>p$ then $a\geq p^2$. $p^2|a^p \implies p^2$ does not divide $a^p - p = b!$ and hence $b<2p$. Now $b!+p \leq (2p-1)! + p < 2 (2p-1)! < 2 \{(2p-1)(1)\}\{(2p-2)(2)\}\cdots \{(p+1)(p-1)\}\{p\} < 2\{p^2\}\{p^2\}\cdots \{p^2\}\{p\} = 2p^{2p-1}< p^{2p} < a^p$ . Which is a contradiction and hence $a = p$. So we have reduced the problem to $$b! = p^p-p$$Claim 3 The above equation has no solution for $p>5$ $$\nu_2(p^p-p) = \nu_2(p^{p-1}-1) = 2\nu_2(p-1)$$For prime $p\geq 7$ the following are distinct $p-1,\frac{p-1}{2} , 4, 2$ and hence their product divides $b!$ $$4(p-1)^2|b! \implies \nu_2(b!) > 2\nu_2(p-1)$$Which is a contradiction. Now we can check that $p=2,3$ indeed gives two solutions with $b=2,4$ respectively. Hence all the solution of the given equation are $$(a,b,p) = (2,2,2) \text{ or } (3,4,3) $$

Suppose $b < p$. If $a \le b$ then taking mod $a$, we get $0 \equiv p + 0 \pmod a$, which means $a \mid p$. This forces $a = p > b$ (because $a > 1$ clearly), a contradiction. On the other hand, if $a>b$ then $a^p \ge (b+1)^p \ge b^p + pb^{p-1} > b! + p$, a contradiction again. Therefore $b \ge p$. We now show that $a = p$. For any prime $q < p$, note that $q \mid b!$ and $q \nmid p$, so $q \nmid a^p$. This means $a$ has no prime divisors smaller than $p$. This means that if $a > p$ then $a$ is in fact at least $p^2$. But then \[b! = a^p - p \ge p^{2p} - p > (2p)!-p \ge (2p-1)!,\]so $b \ge 2p$. This means $p^2 \mid b!$, so $v_p(b! + p) = 1$, a contradiction because $v_p(a^p) = pv_p(a)$ is a multiple of $p$. We have now shown that $a = p$, and we now want to solve for \[b! = p^p - p\]By LTE, \[v_2(p^p-p) = v_2(p^{p-1} - 1^{p-1}) = v_2(p-1) + v_2(p+1) + v_2(p-1) - 1 = 2v_2(p-1) + v_2(p+1) - 1,\]and for $p > 9$, \[v_2(b!) \ge v_2(p!) \ge v_2 \left (4 \cdot \frac{(p-1)}{2} \cdot \frac{(p+1)}{2} \cdot (p-1)\right ) > 2v_2(p-1) + v_2(p+1)-1\]which is a contradiction. Therefore $p = 2,3,5,$ or $7$. A quick check proves that only $p = 2$ and $p=3$ work, and the tuples are $(2,2,2)$ and $(3,4,3)$. EDIT: I don't understand why this is a P5; this is an extraordinarily unremarkable NT where standard methods just work.

Here is My Alternate Small Solution of the Problem : Here $a,b,p$ are positive integers with $p$ prime . $a^p$ is congruent to $a(mod p)$ : $a$ is less than or equal to $p$ . $a^p$=b!+p is congruent to a[p] -> b! is congruent to a[p]. $a^p$=b!+p -->$a^p$ > p --->$a^p$ > p(min) = 2 --> (a not equal to 1) . So $p$ is greater than or equal to p(min) --> $a^p$ is greater than or equal to $a^p(min)$=$a^2$ , p=a --> $p^p$ is congruent to 0[p] -->p($p^p$/p -1) = b!. If $p=a=2$ then $b=2$ . Thus we get $(a,b,p)=(2,2,2)$. Now for $p=a=3$ we will get $b = 4$ . Another we get $(a,b,p)=(3,4,3)$. Now If $p > 3 --> (p+1)! < p(p^p/p -1) < (p+2)!$ and also for all $p$ > $3 -->p(p^p/p -1)$ is not equal to b! .

The solutions are the triples $(2,2,2)$ and $(3,4,4).$ Fistly, if $p=2$ then we have $b! = a^2 - 2$ and since $2$ is not quadratic residue modulo $3$ and $4$ it follows that $b\le 2;$ if $b=2$ then $a^2 = 4$ which produces one of the above solutions, if $b = 1$ then $a^3 = 3$ which does not produce solution. Now we consider the cases where $p$ is odd. Claim: If $p$ is odd and $b\ge p,$ then $(a,b,p) = (3,4,3).$

It remains to deal with the cases where $b< p.$ In this case, we can easily conclude, $$ a^p = b! + p < p^{p-1} + p < p^p$$so $a<p.$ Hence, if $a\le b$ we will have $a \mid a^p - b! \Rightarrow a \mid p,$ this is a contradiciton since $a<p$ and clearly $a$ cannot be $1.$ So $a\ge b+1$ and if we look only at the second and last term of the binomial expansion we will have $$ a^p > {(1+b)}^p \ge b^p + bp > b! + p$$where the last inequality camed from $b<p$. So this case cannot produce any other solutions. The only solutions are hence the ones described in the beginning.

There are two cases: $b < p$ and $b\ge p$. If $b < p$, note that $a\le b$ would imply $a\mid p$. This is absurd unless $a=1$, which would yield the false assertion $1=b!+p > 0! + 2 \ge 3$. Hence $a > b$. Observe that \[a^p - p \ge (b+1)^p - p \ge b^p + pb + 1 - p > b^p \ge b!,\]so this possibility may not occur. If $b\ge p$, note that $p\mid a^p$. Let $a = kp$. As $p$ must divide $a^p$ more than once, $p$ can only divide $b!$ once, hence $b < 2p$. Observe that \[a^p = b! + p \le (2p-1)! + p = p\prod_{k=1}^{p-1} [(p-k)(p+k)] + p = p\left[1 + \prod_{k=1}^{p-1} (p^2 - k^2)\right] \le p\cdot \prod_{k=1}^{p-1} p^2 = p^{2p-1}.\]Thus $a < p^2$, meaning that $k < p$. Remark that if $q$ is a prime dividing $k$, then $q \le k < p \le b$ must divide $b!$ and $a^p$, absurd. Hence $k = 1$ and we seek integer solutions to $p^p - p = b!$. For $p = 2$ and $p=3$ we get the solutions $(2,2,2)$ and $(3,4,3)$. Now assume $p\ge 5$. Remark that $2$ divides $b!$ at least $\lfloor \frac p2\rfloor + \lfloor \frac p4 \rfloor \ge \frac p2$ times, so it must also divide $\frac{p^p-p}{p} = p^{p-1} - 1$ at least $\frac p2$ times. By LTE, then, \[\frac p2 \le \nu_2(p^{p-1} - 1) = \nu_2(p-1) + \nu_2(p-1) + \nu_2(p+1) - 1 < \log_2 (p-1) + \log_2 (p-1) + \log_2 (p+1) = \]\[\log_2 (p-1) + \log_2 (p^2 - 1) \le \log_2 p + \log_2 p^2 = 3\log_2 p.\]Thus $2^{\frac p6} \le p$ by rearranging. Remark that $f(x) = x - 2^{\frac x6}$ has derivative $1-\frac 16 \log_e 2 \cdot 2^{\frac x6}$, so the derivative is negative for $x\ge 6\log_2 (6\log_2 e)$, that is, if $x\ge 19$. As $f(30) < 0$, it is enough to examine the possibilities for $p < 30$. Then $p \in \{5, 7, 11, 13, 17, 19, 23, 29\}$. Observe that $p = 5$ fails because $13\mid 5^2 + 1\mid 5^4 - 1$ cannot divide $b!$. Similarly, $p=7$ fails because $43\mid 7^3 + 1 \mid 7^6 - 1$ cannot divide $b!$. Suppose now that $p\ge 11$. Remark that $2$ divides $b!$ at least \[\lfloor \frac p2\rfloor + \lfloor \frac p4 \rfloor + \lfloor \frac p8\rfloor \ge \frac p2 - \frac 12 + \frac p4 - \frac 34 + 1 \ge \frac{3p}{4} - 1\]times. Then $\frac{3p}{4}-1 \le -1 + 3\log_2 p$ by refining the LTE result. That is, $\frac{p}{4} \le \log_2 p$, so $2^{\frac p4} \le p$. For $p\ge 20$, this is absurd because $\frac{20}{4} \ge 5$, so $2^{\frac p4} \ge 32 > p$. It also is true that $2^{19} > 19^4$ and $2^{17} > 17^4$. Thus we need only consider $p \in \{11, 13\}$. At $p = 11$, \[\nu_2(b!) \ge \lfloor \frac p2\rfloor + \lfloor \frac p4\rfloor + \lfloor \frac p8 \rfloor = 5 + 2 + 1 = 8\]and $\nu_2(p-1)+\nu_2(p-1)+\nu_2(p+1) - 1 = 1 + 1 + 2 - 1 = 3$, so we now need only consider $p=13$. We have $\nu_2(p-1)+\nu_2(p-1)+\nu_2(p+1) - 1 = 2 + 2 + 1 - 1 = 4$, which is still less than $8$, so we are done: the only solutions are $(2,2,2)$ and $(3,4,3)$.

We begin by finding bounds on $a$, $b$: Note that $b < 2p$; otherwise $v_p(b!+p)=1$ and $v_p(a^p) \ge p$ since we would need $p \mid a$. It follows that $a < p^2$. Note that $a \ge p$; otherwise, we note \[b! = a^p-p > a^a > a! \implies b>a \implies a \mid p\]by taking modulo $p$ on our equation, contradiction. It follows that $b \ge p$ as well. Thus we can let $a=xp$, where $1 \leq x < p \leq b$, so \[0 \equiv a^p \equiv b! + p \equiv p \pmod x \implies x=1 \implies a=p.\] Thus our equation becomes $b! = p^p-1 = p(p^{p-1}-1)$. For most values of $p$, we can use Zsigmondy to give us a prime divisor of $p^{p-1}-1$ to be $1 \pmod{p-1}$, forcing it to be $2p-1$. This forces $b=2p-1$, but \[(2p-1)! \equiv p \cdot (p-1)!^2 \equiv p \pmod{p^2},\] contradiction. Hence the only solutions are derived from the Zsigmondy exception cases of $p=2,3$, which give the triplets \[\boxed{(2,2,2), (3,4,3)}. \quad \blacksquare\]

Case 1: $(a,p) \neq 1$. Then $(a,p) = 1$, and $a = kp$ for some positive integer $k$. The equation becomes $(kp)^p = b! + p \iff k^p p^{p-1} = \frac{b!}{p} + 1$. Since $k^p p^{p-1} - 1$ is an integer, so is $\frac{b!}{p}$, and thus $b \geq p$ (since $1,2,\cdots,p-1$ is all coprime to $p$). On the other hand, $b < 2p$. Otherwise, $v_p (b!/p) \geq 1$, which implies $RHS \equiv 1 \pmod p$, a contradiction since $p \mid LHS$. We now show $k=1$. Suppose $k \geq p$. Then \[ (kp)^{p} \geq p^{2p} > (2p-1)! + p \geq b! + p, \]with the second inequality by induction, yielding a contradiction. Hence, $k < p$. Now suppose $q \mid k$ for some prime $q$, then $b < q$, otherwise $RHS \equiv p \not \equiv 0 \pmod q$, a contradiction. Hence no prime divides $k$, and thus $k=1$. The equation now becomes $p^p = b! + p \iff p^p - p = b!$. We count factors of $2$: By LTE Lemma, \[ v_2 (p^p - p) = v_2 (p^{p-1} - 1) = 2 v_2 (p-1) + v_2 (p+1) -1 < 3 \log_2 (p+1) \]and by Legendre's \[ v_2 (b!) = \sum_{i=1}^n \lfloor\frac{b}{2^i} \rfloor \geq \frac{p-1}{2} + \lfloor{\frac{p}{4}}\rfloor. \]for $p \geq 3$. For $p \geq 19$, we have \[ \frac{p-1}{2} + \lfloor{\frac p 4}\rfloor > 3 \log _2 (p+1), \]and so $ 2 \leq p \leq 17$. For $p=17$, $v_2 (p^p - p) =2 v_2 (p-1) + v_2 (p+1) - 1 = 8$ using the formula and $v_2 (b!) \geq 8 + \lfloor \frac{17}{4} \rfloor > 8$, so no solution. The same argument can be made for $p = 11,13$. We now manually check $p = 2,3,5,7$ and find that $(2,2,2)$ and $(3,4,3)$ are solutions. Case 2: $(a,p)=1$. Clearly, $a=1$ has no solution, so exclude it. By FLT, $a^p \equiv a \not \equiv 0 \pmod p$. Thus, $b < p$, otherwise $RHS \equiv 0 \pmod p$. If $a \geq p$, no solution since $a^p - p > (p-1)! \geq b!$. Otherwise, $a < p$ and suppose there is some $q$ such that $q \mid a$, then $b < q$ and $q \leq a < p$. Then $a^p > q^p \geq (q-1)! + p$, contradiction. Hence $a = 1$, which we know has no solution.

PHSH wrote: The solutions are the triples $(2,2,2)$ and $(3,4,3).$ Fistly, if $p=2$ then we have $b! = a^2 - 2$ and since $2$ is not quadratic residue modulo $3$ and $4$ it follows that $b\le 2;$ if $b=2$ then $a^2 = 4$ which produces one of the above solutions, if $b = 1$ then $a^3 = 3$ which does not produce solution. Now we consider the cases where $p$ is odd. Claim: If $p$ is odd and $b\ge p,$ then $(a,b,p) = (3,4,3).$

It remains to deal with the cases where $b< p.$ In this case, we can easily conclude, $$ a^p = b! + p < p^{p-1} + p < p^p$$so $a<p.$ Hence, if $a\le b$ we will have $a \mid a^p - b! \Rightarrow a \mid p,$ this is a contradiciton since $a<p$ and clearly $a$ cannot be $1.$ So $a\ge b+1$ and if we look only at the second and last term of the binomial expansion we will have $$ a^p > {(1+b)}^p \ge b^p + bp > b! + p$$where the last inequality camed from $b<p$. So this case cannot produce any other solutions. The only solutions are hence the ones described in the beginning.

The mistake done was on solutions(3,4,4) but this (3,4,3)

The only solutions are $\boxed{(a,b,p)=(2,2,2),(3,4,3)}$: Case 1: $p=2$ We have that $4\nmid a^2-2\Rightarrow b<4$ which leads to $(2,2,2)$ being the only answer. Now assume that $p$ is odd. Case 2: $p\leq b$ We get $p|a$ so let $a=kp$. Now we get $k^pp^p-p=b!$. Since $p^2\nmid k^pp^p-p$ we must have $b<2p\Rightarrow k<p$. If $k\neq 1$ taking modulo $k$ gives $b<k$, a contradiction. If $k=1$ we get that $p^p-p=b!$. Then $v_2(p^{p}-p)=v_2(p^{p-1}-1)=2v_2(p-1)+v_2(p+1)-1=v_2(b!)$. But $v_2(n!)\leq 2v_2(n)-1$ so $b$ is at most $p+1$. A quick size argument gives the above answer. Case 3: $b<p$ Then we must have $a<p$. Taking mod $a$ we get that $b<a$, a size contradiction.

The number $p$ does not have to be prime; we only need $p \geq 2$. The solution does become more tedious, unfortunately. For convenience, we change the letter $p$ to $c$. That is... Quote: Find all triplets $(a, b, c)$ of positive integers such that $c > 1$ and \[ a^c = b! + c. \] Answer. $(a, b, c) \in \{(2, 2, 2), (3, 4, 3)\}$. Clearly these two triplets work, so now we prove that no other triplets work. Some parts of the solution is taken from the official solution. Note that clearly $a > 1$. Lemma 1. Let $p$ be a prime such that $p \mid a$. Then $\nu_p(c) = \nu_p(b!)$.

Lemma 2. If $b \geq c$, then $(a, b, c) \in \{(2, 2, 2), (3, 4, 3)\}$.

It remains to show that $b \geq c$. We suppose for the sake of contradiction that $b < c$. Lemma 3. We have $a \leq b$ and $a \mid c$.

We now analyze the prime divisors of $a$. Let $p$ be the smallest prime divisor of $a$, and let $k = \nu_p(c)$. First, we show that $k \geq 2$. By Lemma 3, $k \geq 1$. If $k = 1$, then Lemma 1 yields $\nu_p(b!) = 1$, so $b \leq 2p - 1$. By AM-GM inequality, $b! \leq p^{2p - 1}$. On the other hand, $p \leq a < c$ and $p \mid a \mid c$, so $a^c - c \geq p^{2p} - p > p^{2p - 1}$; a contradiction. This shows that $k \geq 2$. Next, we do some estimates. We start by showing that $c$ is a power of $p$. If not, then $c \geq 2p^k$ and \[ b! = a^c - c \geq p^{2p^k} - 2p^k > p^{2p^k - 1}. \]On the other hand, $\nu_p(b!) = k$, so $b \leq p(k + 1) - 1$. Some trivial estimates yield \[ b! \leq (p(k + 1) - 1)! \leq p^p (2p)^p (3p)^p \ldots ((k + 1)p)^p = (k + 1)!^p p^{p(k + 1) - 1}. \]Thus we get \[ p^{2p^k} < (k + 1)!^p p^{p(k + 1)} \iff p^{2p^{k - 1} - (k + 1)} < (k + 1)!. \]Trivially estimating $(k + 1)! \leq p^{0 + 1 + \ldots + k}$ yields \[ 2p^{k - 1} - (k + 1) < \binom{k + 1}{2} \implies 2p^{k - 1} < \binom{k + 2}{2}, \]which is false for $p \geq 3$ and $k \geq 2$. However, if $p = 2$, then $c \geq 3 \cdot 2^k$ and repeating the same argument yields $3 \cdot 2^{k - 1} < \binom{k + 2}{2}$, which is again a contradiction for any $k \geq 2$. Thus, $c$ is a power of $p$. Since $\nu_p(c) = k$, we have $c = p^k$. Since $a \mid c$, now $a$ is also a power of $p$. If $a \neq p$, then $a \geq p^2$ and $a^c - c \geq p^{2p^k} - p^k \geq p^{2p^k - 1}$. Repeating the above process gives a contradiction if $p > 2$, so now assume that $p = 2$. Then $a \geq 4$, $c$ is a power of $2$, and the above estimate gives \[ 2^{2^k - (k + 1)} = 2^{2 \cdot 2^{k - 1} - (k + 1)} < (k + 1)!. \]The estimate $(k + 1)! \leq 2^{0 + 1 + \ldots + k}$ gives $2^k - (k + 1) < \binom{k + 1}{2}$, so $2^k < \binom{k + 2}{2}$. This is false for $k \geq 4$, so $k \leq 3$. Since $a \geq 4$ and $a < c = 2^k$ with $k \leq 3$, then we get $a = 4$ and $c = 8$, which does not work since $4^8 - 8 > 5!$ and $4^8 - 8 \not\equiv 0 \pmod{5}$. Thus, $a = p$ and $c = p^k$ for some $k \geq 2$. Repeating the same procedure yields \[ p^{p^k - 1} < p^{p^k} - p^k = b! < (p(k + 1) - 1)! < (k + 1)!^p p^{p(k + 1) - 1} \implies p^{p^{k - 1} - (k + 1)} < (k + 1)!. \]Again, use the esimate $(k + 1)! \leq p^{0 + 1 + \ldots + k}$. This time, we get \[ p^{k - 1} < \binom{k + 2}{2}, \]which implies either $(p, k) \in \{(5, 2), (3, 2)\}$ or $p = 2$ and $k \leq 6$. Plugging into the previous inequality shows that $(p, k) = (5, 2)$ does not work and $p = 2$ forces $k \leq 4$, so $(p, k) \in \{(3, 2), (2, 2), (2, 3), (2, 4)\}$. That is, $(a, c) \in \{(3, 9), (2, 4), (2, 8), (2, 16)\}$. Working mod $5$ implies that $b < 5$ if $(a, c) \neq (2, 16)$, and direct computation shows that none of the three works. Finall, if $(a, c) = (2, 16)$, then $a^c - c = 65520$, which is not a factorial. Too many repeat of estimations Someone please simplify this proof.

plz help im cooked can anybody read this solution and tell me how correct (and maybe a score out of 7) $$a^p-a=b!+p-a$$So by FLT we get $$a(a^{p-1}-1)\equiv 0 \equiv b!+p-a \mod p.$$Thus, $$b!\equiv a \mod p.$$We may then conclude that if $b>p,$ then $b! \equiv 0 \mod p$. If $b\leq p$, then we know that $$a^p\geq 2^p \geq 2p \geq b!+p$$which shows that the only solution in this case is $b=a=p=2$. So we may now assume that $b! \equiv 0 \mod p$, and consequently $a \equiv 0 \mod p$. Thus $\nu_p(a^p-p)=1$, and thus $\frac{b!}{p}$ cannot be divisible by $p$. Replacing $a$ with $kp$ gives us $$a^p-p=k^pp^p-p=p(k^pp^{p-1}-1)=b!$$Dividing both sides by $p$ and taking mod $p$ gives that the right side must be equal to $(b-p)!\equiv 1 \mod p$. We show that there are no solutions for $p \geq 5$. We first show that $a=p$. If $a=kp$ where $k>1$, then we know that another prime divides $a$, and if it divides $b!$ then we arrive at a contradiction. So this means that $k>2p$. Then we get $a^p>2^pp^2p>b!+p$, so $a=p$. Then $p^p= (2p-1)!+p$. $$p^p-p=(2p-1)!$$The right side is clearly greater than the left side for all values $p\geq 5$, so we only require to test $p=3$. Then we get $(3, 4, 3)$ is another solution. So the only solutions that are possible are $(2, 2, 2)$, and $(3, 4, 3)$. The proof is complete.

Resolving The only solutions are $(2,2,2)$ and $(3,4,3)$, which clearly work. Now we show they are the only ones. Claim: If $p = 2$, $a = b = 2$. Proof: Assume $p = 2$. We have $a^2 = b! + 2$. If $b \ge 4$, then $4\mid b!$, so $a^2 \equiv 2 \pmod 4$, a contradiction. Thus, $b \le 3$ and we can check $b \in \{1,2,3\}$ and get that $a = b = 2$ is the only working solution. $\square$ Henceforth assume $p > 2$. Claim: $b! > p$. Proof: Note that $a = 1$ is obviously not possible. So $b! + p = a^p \ge 2^p > 2p$, meaning $b! > p$. $\square$ Claim: $b \ge p$. Proof: Suppose not. Since $b! > p > 2$, $b \ge3$. Note that \[a^p b! + p \le 2b! < (b+1)! < (b+1)^p, \]so $a < b + 1\implies a \le b$. Thus, $a\mid b!$. Since $a\mid a^p$, $a\mid a^p - b! = p$, so $a \in \{1,p\}$. Since $a \le b < p$, $a < p$, so $a = 1$, absurd. $\square$ Claim: $p\mid a$ and $b< 2p$ Proof: Firstly, \[b \ge p\implies p\mid b! \implies p\mid b! + p = a^p\implies p\mid a\] Now, if $b \ge 2p$, then $p^2 \mid b!$, so $\nu_p(b! + p) = 1$, meaning $\nu_p(a^p) = 1$, contradiction since $p^p \mid a^p$. $\square$ Case 1: $a\ne p$ Then $a > p$. Let $a = xp$ for some $x > 1$. Claim: $x \ge p$. Proof: Let $q$ be a prime divisor of $x$. We show that $q \ge p$, which finishes. Suppose otherwise. Then note that $q\mid b! + p$ but $q \nmid p$ (as $q \ne p$), so $q\nmid b! \implies b < q < p$, absurd. $\square$ We now have $a^p \ge (p^2)^p = p^{2p}$, so $p^{2p} \le b! + p < (b+1)! \le (2p)!$. However, note that \[(2p)! = p \cdot \prod_{i=1}^{p - 1} i(2p - i) \le p \cdot \prod_{i=1}^{p - 1} p^2 = p^{2p - 1} \]by AM-GM, a contradiction. Case 2: $ a=p$ Then $b! = p^p - p = p(p^{p - 1} - 1)$. Note that \[\nu_2(b!) \ge \nu_2((p)!) = p - s_2(p) \ge p - \log_2(p + 1),\]where $s_2$ is the sum of digits in base $2$. Now, by LTE \[\nu_2(p (p^{p - 1} - 1) ) = \nu_2(p^{p - 1} - 1^{p - 1}) = 2 \nu_2(p - 1) < 2 \log_2(p ) \] Combining gives that $p - \log_2(p + 1) < 2 \log_2(p)$, so \[ 2\log_2(p) + \log_2(p + 1) > p\implies \log_2(p^2(p + 1)) > p\] Thus, $2^p < p^2(p + 1)$. Claim: $2^x > x^2(x+1)$ for all integers $x \ge 11$. Proof: Let $f(x) = x^2(x+1)$. We wish to show $2^x > f(x) \forall x \ge 11$. We go by induction on $x$. The base case $x = 11$ is obviously true. . Suppose it's true for $11, 12,\ldots, n$. We have \[ \frac{f(n+1)}{f(n)} = \left( \frac{n+1}{n} \right) ^2 \cdot \frac{n+2}{n+1} = \left( 1 + \frac 1n \right)^2 \cdot \left( 1 + \frac{1}{n+1} \right) \le \frac{12}{11} ^2 \cdot \left( \frac{13}{12} \right) < 2 \] Now, we have \[2^{n+1} = 2^n \cdot 2 > f(n) \cdot 2 > f(n) \cdot \frac{f(n+1)}{f(n)} = f(n+1),\]as desired. $\square$ This immediately implies that $p < 11$, so $p\in \{3,5,7\}$. Note that if $p = 7$, then since $b \ge 7$, $5\mid b! = 7^7 - 7$, which is a contradiction as $7^7 \equiv 3 \pmod 5$. Now if $p = 5$, we have $b! = 5^5 - 5 = 3120$, which isn't a factorial. Therefore, $p = 3$ and $b! = 3^3 - 3 = 24\implies b = 4$. Hence $(a,b,p) = (3,4,3)$, as desired.

storage

If b≥2p no sol By mod p then mod p² If b<p And a>p then no sol As a^p≥ (p+1)^p> (p-1)! +p≥b! +p a≤b Then a|p Umpossible So b<a<p a^p> b! +p So b,a≥p If p= b And a>b umpossible if a<b also impossible By mod p p^p = p! +p might work p^(p-1) > p! (p+1)! > p! +p So together p^p> p! +p For p>2 Which gives 1 sol by bounding 2,2,2 If p<a,b<2p a^p is not divisible by p But b! +p is so umpossible a<p b>p has same problem p<b<2p a≥2p so a=kp (kp)^p = b! +p k=por k≥2p If k≥2p At a min (2p)^p * p^(p) > b! + p So k= p If k= p p^(2p) > b! + p So only one final case remaining a= p p^p = b! +p Take some odd prime q dividing p-1 2(vq(p-1)) = vq(b!) If q||p-1 2= vq(b!)>4 So umpossible q^k||(p-1) 2k = vq(b!) > k + k-1 +k-2 ...... = (k+1)k /2 2> k+1/2 3>k Let k= 2 4 = vq(b!)≥ 2+1 vq(b!) = 3 , 5..... So no 4 Umpossible Thus there is no q So 2^k = p^(p-1) -1 So only p,2 divide b! So b≤4 Now done 3,4,3 Final solution (a,b,c) =(3,4,3),(2,2,2)

The solutions are $(2, 2, 2)$ and $(3, 4, 3)$. If $b \geq 2p$, then $p \mid b! + p$, so $p \mid a^p$, so $p \mid a$. But $\nu_p(a^p) \geq p \geq 2$ and $\nu_p(b! + p) = \min\{\nu_p(b!), \nu_p(p)\} = 1$ since $2 \leq \nu_p(b!) \neq \nu_p(p) = 1$, contradiction. If $p \leq b < 2p$, then once again $p \mid a$. If $a \geq 2p^2$, then \begin{align*} a^p \geq (2p^2)^p = 2^p p^{2p} &= \underbrace{p \cdot p \cdot \dots \cdot p}_{p \text{ times}} \cdot \underbrace{2p \cdot 2p \cdot \dots \cdot 2p}_{p \text{ times}} \\ &> 2 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot 2p \\ &= 2 \cdot (2p)! \\ &> (2p)! + p \\ &> b! + p, \end{align*}contradiction. If $p^2 \leq a < 2p^2$, then let $a = kp$ for an integer $p \leq k < 2p$. Now \begin{align*} b! + p = a^p = (kp)^p = k^p p^p &= \underbrace{p \cdot p \cdot \dots \cdot p}_{p \text{ times}} \cdot \underbrace{k \cdot k \cdot \dots \cdot k}_{p \text{ times}} \\ &> 2 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot k \cdot 1 \cdot 1 \cdot \dots \cdot 1 \\ &= 2k! \\ &\geq k! + p. \end{align*}Therefore, $b! + p > k! + p$, so $b > k$, so $k \mid b!$. Now observe that $k \mid a \mid a^p = b! + p$, so $k \mid p$. Since $p \leq k$, this implies that $k = p$. Hence, $p^{2p} = b! + p$. If $b \geq p + 1$, then taking both sides modulo $p + 1$ yields \begin{align*} (-1)^{2p} &\equiv -1 \pmod{p + 1} \\ 1 &\equiv -1 \pmod{p + 1}, \end{align*}contradiction as $p + 1 \geq 3$. If $b = p$, then we claim $p! + p < p^{2p}$ which yields a contradiction. Indeed, \begin{align*} p^{2p} &= \underbrace{p^2 \cdot p^2 \cdot \dots \cdot p^2}_{p \text{ times}} \\ &> 2 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot p \\ &= 2p! \\ &\geq p! + p, \end{align*}as desired. If $a < p^2$, then let $a = kp$ for an integer $1 \leq k < p \leq b$ so that $k \mid b!$ as above. Now observe that $k \mid a \mid a^p = b! + p$, so $k \mid p$. Since $k < p$, this implies that $k = 1$. Hence, $p^p - p = b!$. If $p = 2$, then $(a, b, p) = (2, 2, 2)$, which works. Otherwise, $p$ is odd. On one hand, \begin{align*}\nu_2(p^p - p) = \nu_2(p^{p - 1} - 1^{p - 1}) = \nu_2(p^2 - 1) + \nu_2\left(\frac{p - 1}{2}\right) \leq \log_2((p^2 - 1)(p - 1)) - 1 < 3 \log_2 p - 1.\end{align*}On the other hand, \[\nu_2(b!) = b - s_2(b) \geq b - (\log_2 b + 1) = \log_2\left(\frac{2^b}{b}\right) - 1.\]Since $\frac{2^x}{x}$ is increasing on $[2, \infty)$, we have \[\log_2\left(\frac{2^b}{b}\right) - 1 \geq \log_2\left(\frac{2^p}{p}\right) - 1 = p - \log_2 p - 1.\]Hence $3 \log_2 p - 1 > p - \log_2 p - 1$, or equivalently, $2^p < p^4$. This fails for all $p \geq 17$, so it suffices to check $p \in \{3, 5, 7, 11, 13\}$. If $p = 3$, then $(a, b, p) = (3, 4, 3)$, which works. If $p = 5$, then $b! = 3120$, bad. If $p = 7$, then $b! = 823536$, bad. If $p = 11$, then $b! = 285311670600$, which is contained between $14! = 87178291200$ and $15! = 1307674368000$, bad. If $p = 13$, then $b! = 302875106592240$, which is contained between $16! = 20922789888000$, $17! = 355687428096000$, bad. If $2 \leq b < p$, then obviously $a < p$ as $p \nmid a$ and if $a \geq p$ then \[b! + p \leq \frac{b}{2} \cdot b^{b - 1} + p \leq b^b < p^p \leq a^p,\]contradiction. If $a < b$, then $a \mid b!$ and $a \mid a^p \mid b! + p$, so $a \mid p$, which implies that $a = 1$. This obviously yields no solutions. If $a \geq b$, then (noting that $a \geq 2$ here) \begin{align*} a^p &= (a - 1) \cdot a^{p - 1} + a^{p - 1} \\ &\geq (a - 1) \cdot \underbrace{a \cdot a \cdot \dots \cdot a}_{p - 1 \text{ times}} + p \\ &\geq 1 \cdot 2 \cdot 3 \cdot 4 \cdot \dots \cdot b \cdot 1 \cdot 1 \cdot \dots \cdot 1 + p \\ &= b! + p. \end{align*}The condition $a = b = 2$, derived from the equality case of the second inequality, is necessary for equality to hold throughout. But this implies that $p = 2$, contradicting the assumption that $b < p$. Lastly, if $b = 1$, then $a^p = p + 1$. Of course, if $a = 1$, then $p = 0$, which is impossible, and if $a \geq 2$, then we can induct on $p$ to get that $a^p > p + 1$. Hence, there are no solutions in this case.

doesn’t feel too tough for n4 We claim that $(a,b,p)=(2,2,2)$ and $(3,4,3)$ are the only solutions. Firstly, we show that $b \le 2p$. Indeed, assume the contrary. Then $p \mid p+ b!=a^p \implies p \mid a \implies p^p \mid a^p$. But $p^2 \nmid p + b!$ since $p^2 \nmid p$, giving us the desired contradiction. Next, we shall prove that $p \mid a$. Indeed, assume otherwise. Then any prime factor $q$ of a must not $b!$ since $q \mid b!+p$ but $q \nmid p$. So $b<q \le a$, But note that $p=a^p - b! >a^p -b^p>p$, which is absurd. Now, note that $a \le b$. If this is not the case then $p=a^p - b! >a^p -b^p>p$, which is contradictory. This shows that $a=p$. Hence, $b!= p^p-p=p(p^{p-1}-1)$, so $b> \ge p$ From here it’s basically same as $(iii)$ of @Leartia.