Phorphyrion wrote: A triple $(a,b,c)$ of positive integers is called strong if the following holds: for each integer $m>1$, the number $a+b+c$ does not divide $a^m+b^m+c^m$. The sum of a strong triple $(a,b,c)$ is defined as $a+b+c$. Prove that there exists an infinite collection of strong triples, the sums of which are all pairwise coprime. My solution: $\bullet $Theorem: For any suffice large prime $p$, $2^{4p}-1$ has a prime divisor $q$ and order of $2$ modulo $q$ is $4p.$ Proof: Trivial by Zsigmondy Theorem, or you can consider the prime divisor of $\frac{2^{4p}-1}{5.\left ( 2^{2p}-1 \right )}.$ This also yields $q \equiv 1 \textrm{(mod } 4p \textrm{)}.$ $\bullet $ Turn back to the problem, for any $n \in Z^+$, i will construct a $n-th$ strong triple from $n-1$ previous triples. Define $(a,b,c)_n$ is the $n-th$ strong triple and $S_n$ is the sum of that triple. With $n=1$, choose $(a,b,c)_1=(2,1,1)$ $\bullet$ Assume i have $n$ strong triples, let $p$ be a prime number which is bigger than $max\left \{ S_1,S_2,...,S_n \right \}$ From Theorem and $CRT$ theorem, exist a prime $q$ congruent to $1$ modulo $4p$ and a triple $(a,b,c)$ satisfied: $$\left\{\begin{matrix}(a,b,c)\equiv \left ( 2;-1;-1 \right ) \textrm{(mod } q \textrm{)} & & \\ (a,b,c)\equiv \left ( 0;0;1 \right ) \textrm{(mod } S_1S_2...S_n \textrm{)} & & \\ a\equiv p \textrm{(mod } p^2\textrm{)} & & \\ b+c\equiv p^2-p \textrm{(mod } p^2\textrm{)} & & \end{matrix}\right.$$ $\bullet$ I will be prove that this triple is strong. Take the contrary, $\exists \, m: a+b+c \, | \, a^m+b^m+c^m \Rightarrow a^m+b^m+c^m \, \vdots \, q \rightarrow 2^m+2(-1)^m \, \vdots \, q$ If $m$ is even, then $2^{m-1}+1 \, \vdots \, q \Rightarrow 2^{2(m-1)}-1 \, \vdots \, q \Rightarrow 2(m-1) \, \vdots \, 4p \rightarrow m-1 \, \vdots \, 2$, a contradiction. So $m$ is odd, $2^{m-1}-1 \, \vdots \, q \Rightarrow m-1 \, \vdots \, 4p \rightarrow (m,p)=1 .$ Hence, $$ \left\{\begin{matrix} a^m \, \vdots \, p^2, v_p(b^m+c^m)=v_p(b+c)+v_p(m)=v_p(b+c)=1 \rightarrow v_p(a^m+b^m+c^m)=1 & & & \\ v_p(a+b+c) \geq 2, \textrm{ because } a+b+c \, \vdots \, p^2. & & & \end{matrix}\right.$$, which is a contradiction. $Q.E.D$

Motivation

Define $f(p)=(p-1)^p+(-p)^p$ for each odd prime $p$. Claim: The set of prime divisors of $f(p)$ is infinite. Proof. Assume FTSOC that $q_1,q_2\ldots,q_k$ are the only prime divisors of $f(p)$. Choose a prime $r\equiv 1\pmod{8\prod{q_i}}$. Notice that $\gcd(f(r),\prod{q_i})=1$, thus $f(r)$ has another prime divisor distinct from $q_1,q_2\ldots,q_k$. Also since $f(r)\equiv 7\pmod{8}$, it has a prime divisor $q\equiv 5,7\pmod{8}$. Now take a sufficiently large $q\equiv 5,7\pmod{8}$ such that $q\mid f(p)$ for some prime $p$. Using Dirichlet, take a large prime $p'\equiv p\pmod{q^2-q}$ and let $(a,b,c)=(1,p'-1,qp'^2-p')$. Since $p'^2\mid\text{LHS}$ we have $p'^2\mid 1+(p'-1)^m$ and so by LTE we have $p'\mid m$. If $m$ is odd we have $q\nmid\text{RHS}$ but $q\mid\text{LHS}$ so the triple is strong. If $m$ is even and $q\mid\text{RHS}$ we have that $q\mid 2x^2+1\mid 4x^2+2$. So $(\frac{-2}{q})=1$, but it's wrong since $q\equiv 5,7\pmod{8}$. Thus in both cases the triple is strong. Notice that $a+b+c=qp'^2$ and since we have chosen $p',q$ large enough, we are done. $\square$