We take all primes ${k}\equiv{1}$ $mod$ $3$ and ${(a, b, c)}\equiv{(k-1, k, k+1)}$, since ${k^{k-1}+(k-1)^{k+1}+(k+1)^k}\equiv{(-1)^{k+1}+1^k}\equiv{0}$ $mod$ $k$. We also have ${k^{k-1}+(k-1)^{k+1}+(k+1)^{k}}\equiv{0}$ $mod$ $3$ due to the different parities of the exponents of the two terms' not divisible by 3. Thus, ${k^{k-1}+(k-1)^{k+1}+(k+1)^k}\equiv{0}$ $mod$ $3k=(k-1)+k+(k+1)$. The problem then reduces to showing that there are an infinite number of primes that are $1$ $mod$ $3$. This has been shown and can be shown using quadratic recopricity along with a proof similar to that of Euclid's proof of the infinitude of primes.$\Box$

The last line also follows by Dirichlet's theorem (more complicated)

Imayormaynotknowcalculus wrote: We take all primes ${k}\equiv{1}$ $mod$ $3$ and ${(a, b, c)}\equiv{(k-1, k, k+1)}$, since ${k^{k-1}+(k-1)^{k+1}+(k+1)^k}\equiv{(-1)^{k+1}+1^k}\equiv{0}$ $mod$ $k$. We also have ${k^{k-1}+(k-1)^{k+1}+(k+1)^{k}}\equiv{0}$ $mod$ $3$ due to the different parities of the exponents of the two terms' not divisible by 3. Thus, ${k^{k-1}+(k-1)^{k+1}+(k+1)^k}\equiv{0}$ $mod$ $3k=(k-1)+k+(k+1)$. The problem then reduces to showing that there are an infinite number of primes that are $1$ $mod$ $3$. This has been shown and can be shown using quadratic recopricity along with a proof similar to that of Euclid's proof of the infinitude of primes.$\Box$ (a,b)=(b,c)=(c,a)=1,so u are wrong.

Here's my solution, which is inspired by post #2's idea. Claim: For any prime $p$, the number $(2p)^{2p} - 1$ has a prime divisor greater than $2p$. Proof: Assume otherwise. Suppose $q$ is a prime divisor of $(2p)^{2p} - 1$. Clearly $q \neq 2, p$, and since $(2p)^{q-1} \equiv 1 \pmod{q}$, we have $(2p)^{gcd(2p, q-1)} = (2p)^2 \equiv 1 \pmod{q}$. Therefore, by LTE, $v_q((2p)^{2p} - 1) = v_q((2p)^2 - 1) + v_q(p) = v_q((2p)^2 - 1)$, which implies $(2p)^{2p} - 1 = (2p)^2 - 1$, a clear contradiction. Hence, the conclusion. Now, we take any prime $p \equiv 1 \pmod{3}$ and a prime divisor $q > 2p$ dividing $(2p)^{2p} - 1$. We claim that $(a, b, c) = (q, q - 2p, q + 2p)$ works. Indeed, notice that $q \neq 2p + 1$ since $3 \mid 2p + 1$ and so $q - 2p > 1$. It is also easy to see that $(a, b) = (a, c) = (b, c) = 1$. Furthermore, \[a^b + b^c + c^a = q^{q - 2p} + (q - 2p)^{q + 2p} + (q + 2p)^q \equiv (-2p)^{2p + 1} + 2p \equiv -2p((2p)^{2p} - 1) \equiv 0 \pmod{q}\]and $a^b + b^c + c^a \equiv a + b + c \equiv 3q \equiv 0 \pmod{3}$ since $a, b, c$ are all odd. Hence, $a + b + c = 3q \mid a^b + b^c + c^a$. We are done since there are infinitely many primes $p \equiv 1 \pmod{3}$, as explained in #2 and #3.

Take $(a,b,c)=(8k+1,8k,8k-1)$ and $k\equiv 1 \ (mod \ 3)$ $a+b+c=24k$ and $a^b+b^c+c^a=(8k+1)^{8k}+(8k)^{8k-1}+(8k-1)^{8k+1}$ $(8k+1)^{8k}+(8k)^{8k-1}+(8k-1)^{8k+1} \equiv 1+0+(-1) \equiv 0 \ ( mod \ 8k)$. Now we have $(8k+1)^{8k}+(8k)^{8k-1}+(8k-1)^{8k+1} \equiv 0+2+1 = 0 \ ( mod \ 3)$

For a prime $p\equiv 1(mod \ 20)$ whose existence can be proven via Drichlet, the triplet $(2p-1,2p+1,p)$ holds the conditions.$\blacksquare$