Baltic Way 2024 · Problem 17

Answer: No. Solution: Assume that there exists a prime $p<d$ such that $p \nmid a b c d$. Then, since $p-1 \mid d!$ and $p \nmid d$, by Fermat's little theorem $d^{d!} \equiv\left(d^{p-1}\right)^{\frac{d!}{p-1}} \equiv 1(\bmod p)$. By the same argument $a^{a!} \equiv b^{b!} \equiv c^{c!} \equiv 1$ $(\bmod p)$, and therefore $a^{a!}+b^{b!}-c^{c!}-d^{d!} \equiv 1+1-1-1 \equiv 0(\bmod p)$. Now we prove that for big enough $d$, the product $P$ of primes less than $d$ is at least $d^{10000}$. Assume $d>2^{\frac{10001 \cdot 10002}{2}}$. Notice that by Bertrand's postulate, the biggest prime less than $d$ is at least $\frac{d}{2}$, the second biggest is at least $\frac{d}{4}$ etc., and 10001-th biggest is at least $\frac{d}{2^{10001}}$. So

$$P \geq \frac{d}{2} \frac{d}{4} \cdots \frac{d}{2^{10001}}=\frac{d^{10001}}{2^{\frac{10001 \cdot 10002}{2}}} \geq d^{10000}$$

Now note that the number of quadruples where $d<2^{\frac{10001 \cdot 10002}{2}}$ is finite, because all the number are bounded above by $d^{2024}$ and hence by $2^{\frac{10001 \cdot 10002}{2}\cdot 2024}$. When $d \geq 2^{\frac{10001 \cdot 10002}{2}}$ we have $a b c d \leq$ $d^{1+3 \cdot 2024}<d^{7000}$ and since $P \geq d^{10000}$, there exist at least two primes $p$ and $q$, less than $d$, that do not divide $a b c d$. But then by our first result, we have $p q \mid a^{a!}+b^{b!}-c^{c!}-d^{d!}$, so it cannot be prime. Remark: The solution can be modified as follows. We can proceed in the first paragraph to conclude that $a^{a!}+b^{b!}-c^{c!}-d^{d!}$ is not prime. Indeed, if $a^{a!}+b^{b!}-c^{c!}-d^{d!}=p$ where $p<d$ then definitely $a>d$ (otherwise $a=b=c=d$ and $a^{a!}+b^{b!}-c^{c!}-d^{d!}=0$ ). Hence

$$\begin{aligned} d & >p=a^{a!}+b^{b!}-c^{c!}-d^{d!} \geq a^{a!}-d^{d!}=\left(a^{(d+1) \cdot \ldots \cdot a}\right)^{d!}-d^{d!} \\ & \geq\left(a^{d+1}\right)^{d!}-d^{d!} \geq a^{d+1}-d>d^{d+1}-d>d^{2}-d=(d-1) d \geq d \end{aligned}$$

contradiction. Then in the last paragraph, there is no need to find two primes less than $d$ that do not divide $a b c d$, one is enough.