We'll do a little more and show the equality case is $(2,2)$ and $(4,5)$. Clearly, if $a = 1$ or $b = 1$ there are no solutions. So, let us assume $a,b > 1$. Now if $a \ge b$, the result follows from $3a \ge 2a + a \ge 2b + 2$, with equality occurring exactly when $(a,b) = (2,2)$. Now, assume that $b > a$, where $b = a+k$ for $k > 0$. Then \[ a! + b! \mid a!b! \implies a! + b! \mid (a!)^2 \]and thus \[ N \overset{\mathrm{def}}{=} 1 + (a+1)(a+2)\dots(a+k) \mid a!. \]Observe that this has no solutions when $a=2$ or $a=3$, and the solution $k=1$ only when $a=4$, corresponding to $(a,b) = (4,5)$. Now, assume that $k \ge \frac{1}{2} a -1 \iff 3a \le 2b+2$, but also $a \ge 5$, hence $k \ge 2$. We will derive a contradiction. Suppose the prime $p \mid N$. Then $p \mid a! \implies p \le a$. But also $p > k$, since if $p \le k$ then $N \equiv 1 \pmod p$. Thus we must have \[ k+1 \le p \le a \]and in particular $p > 2$ since $k \ge 2$. Also, $2p \ge 2k+2 \ge a$. We consider now two cases. First, assume that in fact $2p > a$ for every prime $p \mid N$. In light of this, each prime appears at most once in the prime factorization of $a!$ and we conclude \[ N \le \prod_{k < p \le a} p. \]If $k \ge a$ then $N =1$ which is ludicrous. Otherwise, note that there are $a-k$ numbers in the range, so at most $\left\lceil \frac{1}{2}(a-k) \right\rceil$ primes (since the even numbers are not prime). Then, we get \[ N < a^{\left\lceil \frac{1}{2}(a-k) \right\rceil}. \]On the other hand, $N > a^k$ too. So we derive \[ k < \left\lceil \frac{1}{2}(a-k) \right\rceil \implies k+1 \le \frac{1}{2}(a-k) + \frac{1}{2} \implies k \le \frac 13(a-1). \]But $k \ge \frac{1}{2} a -1$, so this is a contradiction for $a > 4$. In the other case assume $a = 2q$. By inspection assume $q \ge 5$ and note $k = \frac{1}{2} a - 1 = q-1$. In that case we obtain \[ N = 1 + (2q+1)(2q+2) \dots (3q-1) \mid (2q)!. \]Then by the same argument as before we have \[ (2q)^{q-1} < N \le q^2 \prod_{q < p < 2q} p < q^2 \cdot (2q)^{\frac{1}{2}(q -1)} \]since the primes in the product must from from $\{q+2, q+4, \dots, 2q-1\}$. The last inequality rearranges to $(2q)^{q-1} < q^4$ which is clearly false for any prime $q \ge 5$.

For $a \ge b$ then it suffices to check for $b \le a \le 1$. If $a \ge 2$ then it's obvious that $3a \ge 2b+2$. For $a<b$, we have $1+ \frac{b!}{a!} \mid b!= \frac{b!}{a!} \cdot a!$ implies $1+ \frac{b!}{a!} \mid a!$. On the other hand, since there are $b-a+1$ consecutive numbers in the product $\frac{b!}{a!}$ so for any prime $p$ such that $1 \le p \le b-a$ then $p \mid \frac{b!}{a!}$. Therefore, $\gcd \left( (b-a)!, \frac{b!}{a!}+1 \right)=1$. If $b-a+1$ is a composite number then $\gcd \left( (b-a+1)!, \frac{b!}{a!}+1 \right)=1$. This follows $1+ \frac{b!}{a!} \mid \frac{a!}{(b-a+1)!}$, which means $$\frac{a!}{b!}=\underbrace{(a+1) \cdots b}_{b-a \; \text{numbers}} < \underbrace{(b-a+2) \cdots a}_{2a-b-1 \; \text{numbers}}=\frac{a!}{(b-a+1)!}.$$Since $b>a$ so each factor in LHS is greater than the respective factor in RHS, so the only way for the above inequality to be true is $b-a<2a-b-1$ or $3a \ge 2b+2$. If $b-a+1$ is a prime. If $b-a+1 \ne a$ then there always exist a positive integer $k$ such that $a \le k(b-a+1) \le b$. This means $b-a+1 \mid \frac{b!}{a!}$. We do the similar thing like the above case. If $b-a+1=a$ is a prime. From $\gcd \left( (b-a)!, \frac{b!}{a!}+1 \right)=1$ we follow $b-a<2a-b$ or $3a \ge 2b+1$. Since $2a-1=b$ so $a \le 1$. This case can be easily checked. Thus, $3a \ge 2b+2$ as desired.

$b$ cannot be $1$ as $a!+1>a!$. Suppose that $a<\dfrac{2(b+1)}{3}\le b$. The condition is equivalent to $1+(a+1)(a+2)...b|b!$. Let $N=(a+1)(a+2)...b+1$. It's easy to see that the prime factos of $N$ have to be greater than $\dfrac{b}{2}$ and less than or equal to $a$. If $\dfrac{b}{2}\ge a$, this cannot happen, so suppose $\dfrac{b}{2}<a$. As $v_p(b!)\le 1$ for any prime $p>\dfrac{b}{2}$, we infer $$N\le \displaystyle\prod\limits_{\frac{b}{2} <p\le a}p\le a^{a-\frac{b}{2}}$$On the other hand $$N>(a+1)(a+2)...b\ge a^{\frac{b-a}{2}}\cdot \left ( \dfrac{a+b}{2} \right )^{\frac{b-a}{2}}$$Hence $a\le \left ( \dfrac{a+b}{2} \right )^{b-a}<a^{3a-2b}\le a$, a contradiction.

Cute problem, I liked it a lot Observe that $a=b$ means that we want $a \ge 2$ and this clearly follows since $(1.1)$ is not a solution. Also, let $a>b$ and in this case, we assume that $b=1$. But then, $a!+1 \mid a!$ which is false; therefore, $b \ge 2$ and $2b+2 \le 3b <3a$ and the claim holds. Only possibility is $a<b$. Suppose that our result does not hold, that is, $2(b+1)>3a$. We firstly see that $a!+b! \mid a!(a!+b!)-a!b!=(a!)^2$ and thus, $1+(a+1)\cdot(a+2)...(b-1)\cdot b \mid a!$. Now, let $c=b-a$ and $m=(a+1)...(c-1)c$. This translates to $1+m \mid a!$ where $c>\frac{a}{2}-1$. Let $p$ be a prime number with $p \le \frac{a}{2}$ and see that $p$ has a multiple on the range $[a+1,c]$ and so, $p$ does not divide $1+m$. Therefore, the only prime factors of $1+m$ are those in the range $[\frac{a}{2},a]$ and each cannot have a higher exponent than it does in $a!$. Thus, $1+m<(\frac{a}{2}+1)...(a-1)a$ which is false clearly. The claimed result holds.

v_Enhance wrote: We eliminate the case $a = 2p$ by noticing that in this case, $k = p-1$ and we would have $N = 1 + p \cdot (p+1) \dots (2p-1) \equiv 1 \pmod p$. Here,when $a=2p,k=p-1$,$N=1+(2p+1)……(3p-1)\equiv 1+1……(p-1)\equiv 1+(p-1)!\equiv 0(mod p)$ by wilson theorem.And this is a case $3a=2b+2$ holds,maybe I mistook something?

Aiscrim wrote: $b$ cannot be $1$ as $a!+1>a!$. Suppose that $a<\dfrac{2(b+1)}{3}\le b$. The condition is equivalent to $1+(a+1)(a+2)...b|b!$. Let $N=(a+1)(a+2)...b+1$. It's easy to see that the prime factos of $N$ have to be greater than $\dfrac{b}{2}$ and less than or equal to $a$. If $\dfrac{b}{2}\ge a$, this cannot happen, so suppose $\dfrac{b}{2}<a$. As $v_p(b!)\le 1$ for any prime $p>\dfrac{b}{2}$, we infer $$N\le \displaystyle\prod\limits_{\frac{b}{2} <p\le a}p\le a^{a-\frac{b}{2}}$$On the other handl $$N>(a+1)(a+2)...b\ge a^{\frac{b-a}{2}}\cdot \left ( \dfrac{a+b}{2} \right )^{\frac{b-a}{2}}$$Hence $a\le \left ( \dfrac{a+b}{2} \right )^{b-a}<a^{3a-2b}\le a$, a contradiction. why must p be greater than b/2???

Clearly $a, b \ge 2$, so the case where $a \ge b$ is trivial. Let us look when $a<b$. This gives $1+\frac{b!}{a!} | b!$. Assume the contrary and say $3a<2b+2$, or $3a \le 2b+1$. So we are looking at $N=(a+1)(a+2) \cdots b + 1$. The main claim is that all primes that divide $N$ is larger than $\frac{b}{2}$. Let $p|N$. Since $(a+1) \cdots b$ is $b-a$ consecutive numbers multiplied, we must have $p \ge b-a+1$. Now if $\frac{a+1}{2} \le p<\frac{b}{2}$, it is a divisor of $N-1$, so it cannot be a divisor of $N$. Notice that $\frac{a+1}{2} \le b-a+1$. This shows that all primes $b-a+1 \le p<\frac{b}{2}$ cannot be a divisor of $N$. This shows that if $p|N$, we must have $p>\frac{b}{2}$. Also, as $N|b!$, $p \le b$. We can also easily see that $p \le a$, since for all $a+1 \le p \le b$, $p|N-1$. The finish is exactly the same as Aiscrim.

If $a\ge b$ the inequality it's obvious.Now suppose that $a<b$ and let $b=a+x$.So we have to prove that $a\ge 2x+2$ or $x<\left\lfloor\frac{a}{2}\right\rfloor.$ $1+\frac{b!}{a!}\mid b!$,so $1+(a+1)(a+2)\cdot...(a+x)\mid b!=a!\cdot (a+1)(a+2)\cdot...\cdot (a+x)$,following that $$1+(a+1)(a+2)\cdot...(a+x)|a!=x!\cdot\frac{a!}{x!}.$$Since $x!<1+(a+1)(a+2)...(a+x)$ we must have $x!<a!$ or $x<a~(\star)$,so $\frac{a!}{x!}\in\mathbb{N}$. But $x!\mid (a+1)(a+2)\cdot...(a+x)$,so $1+\frac{(a+x)!}{a!}\mid\frac{a!}{x!}$,meaning that $\frac{(a+x)!}{a!}<\frac{a!}{x!}$ or $(a+x)!x!<(a!)^2$. Since the function $(a+x)!x!$ is strictly increasing and we want to prove that $x<\left\lfloor \frac{a}{2}\right\rfloor$,it would suffice to prove that $\left(a+\left\lfloor \frac{a}{2}\right\rfloor\right)!\left(\left\lfloor \frac{a}{2}\right\rfloor\right)!\ge (a!)^2.$ If $a=2k$,then we must prove that $(3k)!k!\ge[(2k)!]^2$,which is equivalent to $(2k+1)(2k+2)...(2k+k)\ge(k+1)(k+2)...(k+k)$.This last inequality is obvious since $2k\ge k$. If $a=2k+1$,then we must prove that $(3k+1)!k!\ge[(2k+1)!]^2.$Let $t_k=\frac{(3k+1)!k!}{[(2k+1)!]^2}.$We have $$\frac{t_{k+1}}{t_{k}}=\frac{(3k+2)(3k+3)(3k+4)(k+1)}{[(2k+2)(2k+3)]^2}> 1,\forall k\ge 1.$$ For $k=4$ the inequality is satisfied and so the inequality is satisfied for any integer $k\ge 4$.But for $k=1,2,3$ the inequality fails!!!!! No problem.For $k=1$ we have $a=3$ and we basically must prove that $x=0$.But form $(\star)$ we have that $x<3$,so $x=0,1,2$. $x=1\implies a=3,b=4\implies 3!+4!\mid 3!4!$ or $1+4\mid 4!$,contradiction! $x=2\implies a=3,b=5\implies 3!+5!\mid 3!5!$ or $1+4\cdot 5\mid 5!$,contradiction! Therefore $x=0$. $k=2$ implies $a=5$ and from $(\star)$ we have that $x<5$. $x=2\implies a=5,b=7\implies 5!+7!|5!7!\implies 1+6\cdot 7|7!$,contradiction! $x=3\implies a=5,b=8\implies 5!+8!|5!8!\implies 1+6\cdot 7\cdot 8|8!$,contradiction! $x=4\implies a=5,b=9\implies 5!+9!|5!9!\implies 1+6\cdot 7\cdot 8\cdot 9|9!$,contradiction! Thus $x=1$,so $a>2x+2$. $k=3$ implies $a=7$ and so $x<7$.Again just some easy cases. This ends our proof.

mathmaths wrote: v_Enhance wrote: We eliminate the case $a = 2p$ by noticing that in this case, $k = p-1$ and we would have $N = 1 + p \cdot (p+1) \dots (2p-1) \equiv 1 \pmod p$. Here,when $a=2p,k=p-1$,$N=1+(2p+1)……(3p-1)\equiv 1+1……(p-1)\equiv 1+(p-1)!\equiv 0(mod p)$ by wilson theorem.And this is a case $3a=2b+2$ holds,maybe I mistook something? You are correct, I messed up. I've fixed the solution (which now needs an extra case). It's surprising how annoying it is to squeeze out the exact equality cases from the shorter solution to the original problem.

Sorry for my bad english A proof which shows the equality cases

Clearly, no solutions exist with $1 \in \{a, b\}$. Also, if $a \ge b$, then $b \ge 2 \implies 3a \ge 3b \ge 2b+2$ automatically. Hence we may assume $2 \le a < b$. Note that \[a! + b! \mid a!(a!+b!) - a!b! = (a!)^2.\]Divide both sides by $a!$, giving \[1 + \frac{b!}{a!} \; \bigg| \; a!.\]Writing $1 + \tfrac{b!}{a!} = 1 + \tbinom{b}{a} (b-a)!$, we see that all prime factors of $1 + \tbinom{b}{a}(b-a)!$ must be greater than $b-a$; hence we can strengthen as \[ 1 + \frac{b!}{a!} \;\bigg|\; \frac{a!}{(b-a)!} \](since, in dividing the right-hand side by $(b-a)!$, we only lose prime factors that are at most $b-a$). Hence \[\frac{b!}{a!} < \frac{a!}{(b-a)!} \implies b!(b-a)! < (a!)^2.\]Note that the left-hand side is increasing in $b$; hence it suffices to prove \[\lfloor \tfrac32a \rfloor! \lfloor \tfrac12a \rfloor! \ge (a!)^2,\]which will establish $b < \lfloor \tfrac32a \rfloor$ and hence $b \le \lfloor \tfrac32a \rfloor - 1 \le \tfrac32a - 1$, as desired. To do so, when $a$ is even simply write \[\begin{aligned} \lfloor \tfrac32a \rfloor! \lfloor \tfrac12a \rfloor!& = \left[ a!\cdot (a+1)(a+2)\ldots \left(\tfrac32a\right) \right] \cdot \left(\tfrac12a \right)! \\ &\ge a! \cdot \left(\tfrac12a+1\right)\left(\tfrac12a+2\right)\ldots(a) \cdot \left(\tfrac12a\right)! \\ &= (a!)^2. \end{aligned}\]When $a$ is odd, this turns out to be a bit of a chore. We'll go by induction. Verify base cases $a= 3, 5, 7, 9, 11$ manually. (Sorry.) Then, suppose $\lfloor \tfrac32k \rfloor! \lfloor \tfrac12k \rfloor! \ge (k!)^2$ for some odd positive integer $k \ge 11$. We have \[\begin{aligned} \lfloor \tfrac32(k+2) \rfloor! \lfloor \tfrac12(k+2) \rfloor! &= \lfloor \tfrac32k \rfloor! \left(\lfloor \tfrac32k + 1 \rfloor \lfloor \tfrac32k + 2 \rfloor\lfloor \tfrac32k + 3 \rfloor\right) \cdot \lfloor \tfrac12k \rfloor! \left(\lfloor \tfrac12k + 1 \rfloor\right) \\ &\ge (k!)^2 \cdot \tfrac{27}{16}k^4 \\ &\ge ((k+2)!)^2 \end{aligned}\]using the inequality $\tfrac{3\sqrt3}{4} k^2 \ge (k+1)(k+2)$, which holds for every $k \ge 11$. This completes the induction.

If $a\geq b,$ we are done unless $b=1$ which is impossible. Hence assume $a<b.$ Also assume $3a\leq 2b+1$ FTSOC. Then $a!+b!=a!\left(1+\tfrac{b!}{a!}\right)\mid a!b!\implies 1+\tfrac{b!}{a!}\mid b!.$ Also $\tfrac{b!}{a!}\mid b!$ and $\gcd(\tfrac{b!}{a!},\tfrac{b!}{a!}+1)=1,$ so we have \[\tfrac{b!}{a!}\left(1+\tfrac{b!}{a!}\right)\mid b!\implies \frac{a!}{1+\tfrac{b!}{a!}}\in\mathbb{Z}\implies 1+\tfrac{b!}{a!}\mid a!.\] Thus every prime divisor of $1+\tfrac{b!}{a!}$ is $\leq a.$ Also since $\tfrac{b!}{a!}$ is a product of $b-a$ consecutive integers, its prime divisors are all $\geq b-a+1\geq \tfrac{a+1}{2}.$ All of these primes can divide $a!$ at most once, so it divides $1+\tfrac{b!}{a!}$ at most once as well. So if $p_1,\cdots, p_k$ are the primes in $[\tfrac{a+1}{2},a],$ we must have $1+\tfrac{b!}{a!}\mid p_1p_2\cdots p_k.$ Then \[1+a^{a/2}<1+(a+1)\cdots (\tfrac{3a-1}{2})\leq1+\tfrac{b!}{a!}\leq p_1p_2\cdots p_k\leq a^k\leq a^{a/2},\]which is a contradiction (modulo roundoff errors for small $a,$ but these can be checked by hand).

Assume the contrary. If a=1 or b=1, b!+1 > b! and a!+1 > a! respectively. Therefore, both a and b are at least 2. If $a \geq b$, the final condition is true, and thus we consider b > a. Since a! * b! is a multiple of a!+b!, b! is a multiple of $1+\frac{b!}{a!}$. A prime factor of the expression $1+\frac{b!}{a!}$ is less than b because b! is a multiple of $1+\frac{b!}{a!}$. Additionally, since $\frac{b!}{a!}$ is a product of b-a numbers a prime factor of $1+\frac{b!}{a!}$ is at least b-a+1. Thus, twice the prime factor is at least 2b-2a+1. Since we assumed the contrary, 2b-2a+1 > a. If twice the prime factor is not greater than b, it is a factor of $\frac{b!}{a!}$. Thus, all prime factors of $1+\frac{b!}{a!}$ are in the range of (b/2, a]. Notice that because of this condition, no square of a prime may divide $1+\frac{b!}{a!}$ because twice any prime factor exceeds b and $1+\frac{b!}{a!}$ divides b!. Since $\frac{b!}{a!}$ is a product of b-a numbers that are each greater than a, $1+\frac{b!}{a!}$ should have at least b-a prime factors. Since they are all in the range of (b/2, a], this means that in the range of (b/2, a], there are at least b-a prime numbers. Since in any six consecutive number, there are at most 2 prime numbers, $\frac{1}{3} * (a-b/2) + 1 > a-b$. Combining this inequality with the contrary to the desired condition, only a small number of cases are left to be solved. ,

Nice problem, pretty simple though... First, as the case when $ a \geq b$ is trivial, let's consider $b=a+k =>$ We need to prove $\frac{a}{2}-1 \geq k$ Hence our condition becomes $a! +(a+k)! \mid a! \cdot (a+k)! => a!(1+(a+1)(a+2)...(a+k)) \mid a! \cdot (a+k)! => 1+(a+1)(a+2)...(a+k) \mid (a+k)! => 1+(a+1)(a+2)...(a+k) \mid a!$ (as $gcd((a+1)...(a+k)+1,(a+1)...(a+k))=1$). Notice that if we can prove that if all primes dividing $a!$ also divides $(a+1)...(a+k)$, we'll be done- but for $k \geq \frac{a}{2}$ and $a$ non-prime this is obvious as all primes lesser than $a$ must be $\leq \frac{a}{2}$, thus we can't have a string of $\frac{a}{2}$ or more numbers if we are to have any hope of satisfying the condition. The case where $a$ is prime is almost identical- just this time we'll get $1+(a+1)(a+2)...(a+k) \mid a$ for $ k \geq \frac{a}2$, which is of course absurd $=>$ Done.

The result trivially holds true if $a=b$. WLOG assume that $b> a$. $$a!+b!\mid a!b!\implies 1+(a+1)(a+2)\ldots b\mid a!$$Suppose $p\mid 1+(a+1)(a+2)\ldots b$, where $p$ is a prime. Then we must have $p\le a$. If $p\le b-a$ then $p$ must divide one of the numbers between $a+1$ and $b$. Thus we must have $a\ge p\ge b-a+1$. Suppose $p=b-a+1$. Suppose $p\nmid (a+1)(a+2)\ldots b\implies a+1\equiv 1 \pmod p\implies b-a+1\mid a$. If $2(b-a+1)\le a\implies 3a\ge 2b+2$. So let's assume that $b-a+1=a$. $$1+(a+1)(a+2)\ldots b\mid a(a-1)\ldots (b-a+1)\implies 1+(a+1)(a+2)\ldots (2a-1)\mid a$$a contradiction. Now let's assume that $p\mid (a+1)(a+2)\ldots b$. Then we will have $$1+(a+1)(a+2)\ldots b\mid a(a-1)\ldots (b-a)\implies (a+1)(a+2)\ldots b<a(a-1)\ldots (b-a) \implies 2b+1<3a\implies 3a\ge 2b+2\quad\blacksquare$$

Cute Okay my solution is quite identical to others so I'll hide.

Note that if $a=1,$ then $b!+1\mid b!,$ a clear contradiction and similarly we can$'$t have $b=1,$ so in fact $a, b\geq 2.$ Now if $a\geq b,$ then obviously $3a\geq 2b+a\geq 2b+2,$ so assume $a\leq b-1.$ Next observe that $$a!(1+b(b-1)\cdots (a+1))\mid a!b!$$$$\iff 1+b(b-1)\cdots (a+1)\mid b!=b(b-1)\cdots (a+1)a!$$ Since $\gcd\bigg(1+b(b-1)\cdots (a+1), b(b-1)\cdots (a+1)\bigg)=1,$ it follows that $$1+b(b-1)\cdots (a+1)\mid a!$$ Let $p$ be an arbitrary prime factor of $1+b(b-1)\cdots (a+1).$ By the divisibility above$,$ $p\leq a.$ Let $k$ denote the unique positive integer for which $(k-1)p\leq a<kp.$ If $kp\leq b,$ then $kp\mid b(b-1)\cdots (a+1)$ and $p\mid 1+b(b-1)\cdots (a+1),$ contradiction$.$ Therefore $kp> b$ and consequently $$\hspace{1cm}a+p\geq kp\geq b+1\hspace{1cm} (*)$$$$\Rightarrow p\geq b-a+1=:c+1\hspace{1cm}$$and that is true for every prime factor of $1+b(b-1)\cdots (a+1).$ This yields $$\gcd\bigg(1+b(b-1)\cdots (a+1), c!\bigg)=1$$and as $a\geq p\geq c+1,$ we obtain $$1+b(b-1)\cdots (a+1)\mid a(a-1)\cdots (c+1)\hspace{1cm}(\square)$$Making quick estimates$,$ $$a^{b-a}<1+\underbrace{b(b-1)\cdots (a+1)}_{b-a \; \text{numbers}}\leq \underbrace{a(a-1)\cdots (c+1)}_{a-c \; \text{numbers}}\leq a^{a-c}$$and we get $b-a<a-c \iff 2b+1\leq 3a.$ Assume that the equality is achieved$.$ Now return to $(*).$ If all the equalities there are also achieved$,$ then $a+p=kp=b+1,$ so for some prime $p$ we have $p\mid a, \mid b+1,$ which contradicts $3a=2b+1.$ Therefore one of these inequalities is strict and we actually see that $p>b-a+1=c+1$ for every prime factor $p$ of $1+b(b-1)\cdots (a+1).$ Lastly$,$ in the divisibility $(\square)$ we can drop the $(c+1) \hspace{0.1cm}$ term on the right due to the previous paragraph$.$ This time our sloppy estimate gives $$a^{b-a}<1+\underbrace{b(b-1)\cdots (a+1)}_{b-a \; \text{numbers}}\leq \underbrace{a(a-1)\cdots (c+2)}_{a-c-1 \; \text{numbers}}\leq a^{a-c-1}$$and we finally obtain $b-a<a-c-1\iff 2b+2\leq 3a$ as desired$.$ $\hspace{1cm}\blacksquare$

Solved with L567.

$b=a$ trivially holds for $a,b\ge 2$ and neither can be equal to 1, so we can WLOG assume $b>a\ge 2$ from here onwards. We have that $1+(a+1)(a+2)\ldots b$ is a factor of $b!$. Since it is relatively prime to $\frac{b!}{a!}$, it must be a factor of $a!$ as well. If $b\ge \frac{3a-1}{2}$, $b-a>\frac{a}{2}$ so $1+(a+1)(a+2)\ldots b$ will be relatively prime to all positive integers between $1$ and $\frac{a}{2}$ inclusive. However, since prime factors of $a!$ greater than $\frac{a}{2}$ only divide $a!$ once, we need $1+(a+1)(a+2)\ldots b\le \frac{a!}{\left\lfloor\frac{a}{2}\right\rfloor !}$ which is absurd.

Suppose for contradiction that $3a < 2b+2.$ Note that $a\ge b$ is trivial, so consider $a<b.$ Dividing by $a!,$ it must be true that $$1+b(b-1)\cdots(a+1)\mid b!.$$Additionally, since $\gcd(b!/a!, b!/a!+1) = 1$ and $b!/a! \mid b!,$ $$\frac{b!}{a!} \left(1+\frac{b!}{a!}\right) \mid b! \iff c= 1 + b(b-1)\cdots(a+1) \mid a!.$$Claim: All prime divisors $p$ of $c$ must satisfy $a/2 < p \le a.$ Proof. Note by assumption we have $b-a \ge \frac{a-1}{2}$ and so for all $p\le \frac{a-1}{2},$ there exists a multiple of $p$ in $[a+1, b]$ and if $a$ even then for $p=a/2,$ we have $3a/2 <b+1.$ $\blacksquare$ Moreover, since $c\mid a!$ and each $v_p(a!) = 1$ for $a/2 < p \le a,$ then this implies $$c = 1 + \underbrace{b(b-1)\cdots(a+1)}_{\ge \frac{a-1}{2}} \le \prod_{a/2 < p \le a} p < \underbrace{a(a-1)\cdots ((a+1)/2)}_{\le \frac{a-1}{2}},$$Absurd. $\blacksquare$

I think my solution is different from others, so I post it. $\bullet$ If $a=b$ and $a=1$ we get contradiction , so $ 3a\geq 2b+2$ for $a=b$. $\bullet$ If $a>b$ then $3a\geq 3b+3>2b+2$. $\bullet$ If $b>a$ then let $b=a+t$. Then we need to prove $a\geq 2t+2$. Assume $a<2t+2$. $a!+(a+t)! \mid a!\cdot (a+t)!$ Let $a!\cdot N=(a+t)!$ Then $1+N\mid a!\cdot N$. Since $\gcd(N,N+1)=1$, $1+N\mid a!$ Obviously, if $t>a$ then $1+N>a!$ and we contradiction, so $t<a$ $\implies$ $t!\mid a!$. Since $N$ is multiple of $t$ consecutive numbers $t!\mid N$ $\implies$ $\gcd(N+1,t!)=1$ $\implies$ $t!(N+1)\mid a!$ Now I will prove that $t!(N+1)>a!$ where $a<2t+2$. $Case\ 1:$ $a$ is even. Let $a=2m$. Then $t\geq m$. We need to prove $t!\cdot \frac{(2m+t)!}{(2m)!}+t!>(2m)!$ For minimality of LHS take $t=m$. And then we get much more poverful in equality that $m!\cdot \frac{(3m)!}{(2m)!}>(2m)!$ which true because $(2m+1)\cdots (3m)>(m+1)\cdots (2m)$ $Case\ 2:$ $a$ is odd. Let $a=2m+1$. Then $t\geq m$ and we want to prove $t!\cdot \frac{(2m+1+t)!}{(2m+1)!}+t!>(2m+1)!$ If $t\geq m+1$ then after taking $t=m+1$ similarly we can prove that $t!\cdot \frac{(2m+1+t)!}{(2m+1)!}>(2m+1)!$ So there remains just $1$ case $t=m$. In this case I won't write inequality directly, I will prove that $t!\cdot (N+1)\nmid a!$ Assume that $t!\cdot (N+1)\mid a!$. It means $(N+1)\mid \frac{a!}{t!}$ or $A=\frac{(3m+1)!}{(2m+1)!} +1\mid \frac{(2m+1)!}{m!}=B$. We have that $m!\mid B$ and $m!\mid A$.So $\gcd(A+1,m!)=1$ $\implies$ $m!\cdot (A+1)\mid B$. Since $\frac{(3m+1)!}{(2m+1)!}\cdot m!>\frac{(2m+1)!}{m!}$ (It's obvious, so I didn't write proof of it.) we get $A\cdot m!>B$ which is contradiction, so we are done

Suppose FTSOC that $b\ge \tfrac32a-1$. WLOG $a\le b$, so \[ N:=1+b(b-1)(b-2)\cdots (a+1) \mid b!. \]Consider a prime factor $p\mid N$. Claim 1: $p\in (b-a,a]$, and $p\nmid b,b-1,\ldots,a+1$. Proof: We cannot have $p \in (b,\infty)$ since then $p\nmid b!$. We cannot have $p\in (a,b]$ since then $p\mid b!$ but $p\mid N-1$, so in particular $p\nmid N$. We cannot have $p\in [1,b-a]$ since then $p\mid b!$ but $p$ divides at least one of $\{a+1,\ldots,b\}$, so $p\mid N-1$, so $p\nmid N$. Therefore, we must have $p\in (b-a,a]$. (Note this implies $b\le 2a$.) In fact, we can further say that $p\nmid b,b-1,\ldots,a+1$. $\blacksquare$ Claim 2: $\nu_p(N) = 1$. Proof: We have $p\ge b-a+1>\tfrac12a$ since $b>\tfrac32a-1$. Specifically, if $a$ is even, then $p\ge \tfrac12a+1$, and if $a$ is odd, then $p\ge \tfrac12a+\tfrac12$. In both cases, $2p \ge a+1$. Assume FTSOC $p^2\mid N$. Then $p^2\mid b!$. But recall $p\nmid b,b-1,\ldots,a+1$, so we must have $p^2\mid a!$. But then $2p\le a$, contradiction. $\blacksquare$ Combining the above two claims along with the very weak statement that primes are odd implies $N\le a(a-2)(a-4)\cdots (b-a+2)$ if $a$ is odd and $N\le (a-1)(a-3)(a-5)\cdots (b-a+2)$ if $a$ is even. These are both much less than $1+b(b-1)\cdots (a+1)$ (almost a square root of this), so we are done.

ISL marabot solve Suppose FTSOC that $3a<2b+2$. If $a>b$, then the maximum value for $2b+2$ is $2(a-1)+2=2a$, which is less than $3a$. If $a=b$, then $3a<2a+2\implies a=1$, which is not a solution. So $a<b$. Let $b=a+c$ So \[a!+b!=a!(1+(a+1)(a+2)\cdots (a+c))\mid a!b!\implies 1+(a+1)(a+2)\cdots (a+c)\mid b!\] Let $n=1+(a+1)(a+2)\cdots (a+c)$. Since $n$ is relatively prime to all prime factors in $b!$ that are not in $a!$, $n\mid a!$. Let $p$ be a prime dividing $n$. Since $p\mid a!$, $p\le a$. Note that no prime less than or equal to $c$ divides $n$, because that prime will divide $(a+1)(a+2)\cdots (a+c) $ So $c<p\le a$. Since $3a<2b+2$, we have $2b>3a-2\implies b>1.5a-1$. Now $p\ge c+1>0.5a$, so $2p>a$. This implies $\nu_p(n)=1$. For odd $a$, we have $n\le a(a-2)(a-4)\cdots (c+1)$ and if $a$ is even, we have $n\le (a-1)(a-3)\cdots (c+1)$. Both of these are less than $1+(a+1)(a+2)\cdots(a+c)$, a contradiction.

Here's a different (but long) proof by using $1 + \frac{b!}{a!} \le \gcd \left( 1 + \frac{b!}{a!}, a! \right)$. We will also more strongly shoe that only equality cases are $(a,b) = (2,2),(4,5)$. Case $a=b$ is easy. Assume $a \ne b$. Observe $a,b \ge 3$. If $a > b$ we are done (and no equality case). Now assume $a < b$. Observe, \begin{align*} a! + b! &\mid a!b! \implies a! + b! \mid (a!)^2 \implies 1 + \frac{b!}{a!} \mid a! \\ &\implies 1 + \frac{b!}{a!} \le \gcd \left( 1 + \frac{b!}{a!}, a! \right) \qquad \qquad (1) \end{align*}Note from $b! < (a!)^2$ we also have $b < 2a$. Observe $(b-a)! \mid \frac{b!}{a!}$. Thus $\gcd \left(1 + \frac{b!}{a!} , (b-a)! \right) = 1$. Hence $$ \gcd \left( 1 + \frac{b!}{a!}, a! \right) \le \gcd \left( 1 + \frac{b!}{a!}, \frac{a!}{\gcd(a!,((b-a)!)^{\infty})} \right) \textcolor{blue}{\le} \gcd \left( 1 + \frac{b!}{a!} , \frac{a!}{(b-a)!} \right) \le \frac{a!}{(b-a)!} \qquad \qquad (2)$$Combining this with $(1)$ we obtain the inequality $$ \frac{b!}{a!} < \frac{a!}{(b-a)!} \qquad \qquad (3) $$Now as $ b > a$, so we have that $$ \frac{b!}{a!} \ge \frac{ (b - (b-a))!}{(a - (b-a))!} = \frac{a!}{(2a- b)!} $$So it follows that \begin{align*} 2a - b > b-a \implies 3a > 2b \\ \implies 3a \ge 2b + 1 \qquad \qquad (4) \end{align*}Since we now know $a \ge 2(b-a)$, so the blue inequality ($\textcolor{blue}{\le}$) in $(2)$ can now be strengthened to $$ \gcd \left(1 + \frac{b!}{a!} , \frac{a!}{\gcd( a!,((b-a)!)^\infty)} \right) \le \gcd \left( 1 + \frac{b!}{a!}, \frac{a!}{(b-a) \cdot (b-a)! } \right) \le \frac{a!}{(b-a)!} \cdot \frac{1}{(b-a)} $$So we have the inequality $$ \frac{b!}{a!} < \frac{a!}{(b-a)!} \cdot \frac{1}{(b-a)} \qquad \qquad (4) $$Now we consider some cases (according to when $(4)$ can be satisfied and also $3a \le 2b+2$). CASE 1: $a=2k+1$ and $b = 3k+1$. We will show $(4)$ does not hold true. Observe $(4)$ is equivalent to \begin{align*} \frac{(3k+1)!}{(2k+1)!} &< \frac{(2k+1)!}{k!} \cdot \frac{1}{k} \\ \iff (3k+1)(3k) \cdots (2k+2) &< (2k+1)(2k) \cdots (k+1) \cdot \frac{1}{k} \\ &= (2k+1)(2k-1)(2k-2) \cdots (k+2)(2k + 2) \end{align*}Above is clearly not true as LHS terms are larger than RHS (both have $k$ terms). CASE 2: $a=2k+2$ and $b = 3k+2$. Now for $k=1$ we indeed have $4! + 5! \mid 4! \cdot 5!$. We now show no solutions for $k \ge 2$. The blue inequality in $(2)$ can be further strengthened to $$ \frac{a!}{\gcd(a!,((b-a)!)^\infty)} \le \frac{(2k+1)!}{k!} \cdot \frac{1}{2k} \cdot \frac{2k+2}{2} = (2k+1)(2k-1)(2k-2) \cdots (k+1)(k+1) $$Thus we obtain \begin{align*} (2k+1)(2k-1)(2k-2) \cdots (k+1)(k+1) &> \frac{b!}{a!} = \frac{(3k+2)!}{(2k+2)!} = (3k+2) \cdots (2k+4)(2k+3) \end{align*}Which implies \begin{align*} k+1 > \left( \frac{3k+2}{2k+1} \cdot \frac{3k+1}{2k-1} \cdots \frac{2k+5}{k+3} \right) \cdot \frac{2k+4}{k+2} \cdot \frac{2k+3}{k+1} > \left( \frac{3}{2} \right)^{k-2} \cdot 4 \end{align*}But by induction on $k \ge 2$, it can be proven this inequality is always false. So we obtain our desired contradiction, which completes the proof. $\blacksquare$

Note that if $b<a$, it is trivial, so suppose $a\ge b$. Let $b-a = c$, and suppose for contradiction that $2c+1\ge a$. Then, \[1+ (a+1)\dots (a+c) \mid b!\]but every factor of the numerator not bigger than $c$ or bigger than $a$ divides the denominator-1, so \[1+(a+1)\dots (a+c) \mid (c+1)\dots (a).\]Therefore, \[1+a^c\le 1+(a+1)\dots (a+c) \le (c+1)\dots (a)<a^{a-c}\]so $a-c>c$, so $a = 2c+1$. \[1+(2c+2)\dots (3c+1) \mid (c+1)\dots (2c+1)\]but $c+1$ is prime with the denominator, hence \[1+(2c+2)\dots (3c+1) \le (c+2)\dots (2c+1)\]\[\implies 1+(2c+2)^c \le (2c+1)^c,\]a contradiction.

Clearly $a,b = 1$ fails and clearly this question is true if $a \geq b$. Thus, assume $2 < a < b$. Furthermore, write $b = a+k$, and assume for the sake of contradiction that $2b + 2 \geq 3a \Rightarrow b \geq \left\lfloor\frac{a}{2} \right\rfloor + a$ (it is not hard to show that the smallest positive integer greater than $\frac{3a}{2} - 1$ is $a + \left\lfloor\frac{a}{2} \right\rfloor$.) First, we rewrite: \[ a! + b! \mid a!b! \Rightarrow a! + b! \mid (a!)^2 \Rightarrow \left. a!\left(1 + \prod\limits_{i=1}^{k} (a+i)\right) \right| a!^2 \Rightarrow \left. \left(1 + \prod\limits_{i=1}^{k} (a+i) \right) \right| a! \]Now, consider a prime $p \leq k$. By looking $\pmod{p}$, it is easy to see that $p \left| \prod\limits_{i=1}^{k} (a+i) \right.$, which implies that it does not divide $ \prod\limits_{i=1}^{k} (a+i) + 1$. Therefore, we can in fact write that \[ \left.\left(1+ \prod\limits_{i=1}^{k} (a+i) \right) \right| \prod\limits_{i=k+1}^{a} i \]because any term on the right with prime factor less than $k$ (i.e., all numbers less than or equal to $k$) are discarded. Note that if $k \geq a-k \Rightarrow k \geq \frac{a}{2}$, then we immediately reach a contradiction because there are at least as many terms on the left and the right and each term on the left is larger. The only time this is not true is when $k$ is exactly $\left\lfloor\frac{a}{2} \right\rfloor$ and $a$ is odd; however, in this case, note that at least one of $k+1$ and $k+2$ is divisible by $2$ and therefore has all prime factors $\leq k$, and a similar argument can be repeated (except when $k = 1$, implying $a = 3, b = 4$, which can be checked by hand to fail).

There are no solutions when $a$ or $b$ is $1.$ If $a\ge b$ we are already done. Thus, let $b>a$, so we have $1+b(b-1)\dots (a+1)\mid b!$ Assume $3a< 2b+2$ then $b-a>\frac{a}{2}$. Note that $b(b-1)\dots (a+1)+1$ is relatively prime to $b(b-1)\dots(a+1)+1$ which implies that $b(b-1)\dots(a+1)+1\mid a!.$ For any prime $p\le \frac{a}{2}$, there are at least $p$ terms in $b(b-1)\dots(a+1)$ which means one of them is divisible by it, which means $p\nmid a!$. Just taking $p$ will give $a<4$. $~$ Taking $a=2$, we have $2+b!\mid 2b!$ which implies $b!=2$, which satisfies what we are trying to prove. Taking $a=3$ we have $6+b!\mid 6b!$ which implies $b!=3,6,12,30$ but only $b!=6$ makes sense, but this also satisfies what we are trying to prove, so we done.

Assume the contrary that $3a\leq 2b+1$ so $b-a\geq \frac{a-1}{2}$. Obviously, $b\geq 2$. Otherwise, we have $a!+1\mid a!$ which is absurd. Thus, if $a\geq b$ then $3a\geq 3b\geq 2b+2$, a contradiction. Now, suppose that $a<b$. \begin{align*} a!(1+(a+1)(a+2)\dots (b))=a!+b!\mid a!b! &\implies 1+(a+1)(a+2)\dots (b)\mid b! \\ &\implies 1+ (a+1)(a+2)\dots (b)\mid a!\cdot ((a+1)(a+2)\dots (b)) \\ &\overset{\text{gcd}}{\implies} 1+(a+1)(a+2)\dots (b)\mid a! \end{align*}Let $p$ be a prime factor of $1+(a+1)(a+2)\dots (b)$. Then $p$ also divide $a!$ so $p\leq a$. We claim that $p>\frac{a}{2}$. Assume the contrary that $p\leq \frac{a}{2}$ so $p\leq \lfloor a/2\rfloor$. Since $b-a\geq \frac{a-1}{2}\implies b-a\geq \lfloor a/2\rfloor$, must be a multiple of $p$ in the at least $\lfloor a/2 \rfloor$ numbers: $a+1,a+2,\dots,b$ so $p\mid 1+(a+1)(a+2)\dots(b)$ which is impossible. Thus, we have $a\geq p>\frac{a}{2}$. Recall that $p\mid 1+(a+1)(a+2)\dots (b)\mid a!$ so $v_p(1+(a+1)(a+2)\dots (b))\leq v_p(a!)=1$. Hence, we have $$1+(a+1)(a+2)\dots (b)= p_1p_2\dots p_k\leq \prod_{\text{All primes }p\in \left[(a+1)/2,a\right]}p\leq ((a+1)/2)((a+1)/2+1)\dots (a)$$which is a clear contradiction. (One could easily check some small annoying case like $a=2,3$.)

A beatiful Number Theory...My solution looks like others , So I will not post full solution... Main Idea

shinichiman wrote: On the other hand, since there are $b-a+1$ consecutive numbers in the product $\frac{b!}{a!}$ No $\frac{b!}{a!}=b(b-1)...(a+1)$ so there are only $b-a$ consecutive numbers.

This is probably wrong. Anyway, FTSOC $2b+1 \ge 3a$. Then $b>a$, then the problem hypothesis translates to $\frac{b!}{a!}+1 | b!$. Take a prime divisor $p$ of $\frac{b!}{a!}+1$, then $p \ge b-a+1$, otherwise $p|\frac{b!}{a!}$. Now, $p$ divides $b!$ and $a!+b!$ so it divides $a!$, and thus $p\le a$, and $2p \ge 2(b-a)+2 \ge a+1 >a$ where we used $2b+1 \ge 3a$. Thus, $v_p(a!)=1$. But $v_p(\frac{b!}{a!}+1)\ge1$, thus $v_p(b!+a!) \ge 2$, which implies $v_p(b!)=1$, and using $\frac{b!}{a!}+1 | b!$, we get that $v_p(\frac{b!}{a!}+1)=1$. Thus $\frac{b!}{a!}+1$ is the product of some distinct primes in the interval $(\frac{a}2;a]$. This gives us the bound $\frac{b!}{a!}+1\le a^{\frac{a+1}4}$, and using $2b+1 \ge 3a$, we get that $\frac{(\lceil \frac{3a-1}{2} \rceil !)}{a!}+1\le a^{\frac{a+1}4}$, and this is actually impossible.

ZNatox wrote: This gives us the bound $\frac{b!}{a!}+1\le a^{\frac{a+1}4}$. How ?

PNT wrote: ZNatox wrote: This gives us the bound $\frac{b!}{a!}+1\le a^{\frac{a+1}4}$. How ? Ok, I messed it up a bit, but fortunately it's salvageable. $\frac{b!}{a!}+1$ is the product of some odd primes in the interval $(\frac{a}2;a]$, but this interval contains at most $\frac{a-\lceil a/2 \rceil +1}2 +1 \le \frac{a+6}4$ odd numbers. And so: $\frac{b!}{a!}+1=\prod_{p\in S \subset \mathbb P} p \le a^{|S|} \le a^{\frac{a+6}4}.$ This fails for $a\ge 7$.

We may assume that $a \leq b$, otherwise the result is obvious. Note that we may optimize the condition to read $N = \frac{b!}{a!} + 1$ divides $b!$. As $N$ is relatively prime to $\frac{b!}{a!}$, $N$ must in facct divide $a!$. Now suppose that $3a < 2b+2$ or $b > \frac{3a-2}2$. This implies that there are at least $\frac{a-1}2$ factors in $\frac{b!}{a!}$. In particular, any prime at most $\frac{a-1}2$ cannot divide $N$; as a result, $N$ consists of primes $p \in \left[\frac{a-1}2, a\right]$. On the other hand, for any such prime $p$, $p$ appears exactly once in $a!$. Thus, we have $$a^{(a-1)/2} < \frac{b!}{a!} + 1 \leq \prod_{a/2 \leq p \leq a} p < a^{(a+3)/4}.$$This is impossible for $a \geq 5$, so checking small cases reveals two solutions $(2, 2)$ and $(4, 5)$, both of which work.