Do you have any other conditions for example ab=cd

Com10atorics wrote: Prove that for any natural numbers $a,b,c$ and $d$ there exist infinetly natural numbers $n$ such that $a^n+b^n+c^n+d^n$ is composite. For $a=b=c=d=1$ the result is obvious. We can assume that $a^n+b^n+c^n+d^n$ is strictly increasing. Assume otherwise. Then we have an $N$ satisfying that for any $n\geq N$ the number $p=a^n+b^n+c^n+d^n$ is prime. Since \[ a^{np}+b^{np}+c^{np}+d^{np}\equiv a^n+b^n+c^n+d^n\equiv0\pmod{p} \]and $a^{np}+b^{np}+c^{np}+d^{np}>p$ this is false.

The case $a=b=c=d=1$ is trivial. In the other cases: Let $p$ be a prime divisor of $(a+b+c+d)$. If $gcd(a,p)=1$ then based on the little Fermat theorem $a^{p-1}\equiv 1\pmod p$ and consequently for every $k \in N$: $$a^{k(p-1)}\equiv 1\pmod p$$On the other hand if $gcd(a,p)>1$, then $p$ is a divisor of $a$. So we can conclude that $$a^{k(p-1)+1}-a\equiv a(a^{k(p-1)}-1) \equiv 0 \pmod p$$Similarly we can conlcude that for any $k \in N$: $$b^{k(p-1)+1}-b \equiv 0 \pmod p$$and $$c^{k(p-1)+1}-c\equiv 0 \pmod p$$and $$d^{k(p-1)+1}-d \equiv 0 \pmod p$$Summing up these results we obtain that for every $k \in N$: $$a^{k(p-1)+1}+b^{k(p-1)+1}+c^{k(p-1)+1}+d^{k(p-1)+1} - (a+b+c+d) \equiv 0 \pmod p$$But $p$ is a divisor of $(a+b+c+d)$, thus it is a divisior of $a^{k(p-1)+1}+b^{k(p-1)+1}+c^{k(p-1)+1}+d^{k(p-1)+1}$ too. Thus for $ n = k(p-1) +1 $ (where $k$ is any natural number) the given number is composite.