# Infinitude of primes of the form $a^n+b$

I was playing with the form $a^n+b$ such that $a,b,n \in \mathbb{N} \geq 1$, and i was changing the values of $a,b$ form $(1,1)$ up to $(10,10)$ the amazing thing i found is :

If the form $a^n+b$ have at least two primes for two values of $n$ then its a necessary and sufficient condition for the form $a^n+b$ to have infinitely many primes

Is this true? have any one dealt with such condition? could you give a counter-example meaning fixed $a,b$ and the form $a^n+b$ have at least $2$ primes yet there are no infinitely many primes of the form $a^n+b$.

In my work the candidates according to my criteria to have infinitely many primes are :

For $a=1$ the values for b are $b=1,2,4,6$ which is obvious since $2,3,5,7$ are primes.

For $a=2$ the values for b are $b=1,3,5,7,9$

For $a=3$ the values for b are $b=2,4,8$

For $a=4$ the values for b are $b=1,3,7,9$

For $a=5$ the values for b are $b=2,4,6,8$

Now if the criteria is true then this would have immense consequences on when does the form $a^n+b$ contains infinitely many primes.

Granted that some of the fixed $a,b$ will not yields a primes right away yet they might have infinitely many primes, but as far as i know all the fixed numbers $a,b$ which are important for mathematicians contains primes for very small numbers $n$.

Thanks to you all.

• The simplest non-trivial instance here, $n^2+1$, is already unknown. That is one of Landau's Problems ....he put it in the class with Twin Primes and Goldbach. – lulu Sep 4 '17 at 11:02
• For $a=2$, $b=1$, we know that $a^n+b$ is prime for $n=1,2,4,8,16$, but no further $n$ for which $2^n+1$ is prime is knwon. – Hagen von Eitzen Sep 4 '17 at 11:03
• @lulu i am talking about $a^n+b$ and not $n^a+b$ . – Ahmad Sep 4 '17 at 11:06
• Ah, sorry. Misread. But, as has been been pointed out, little is known about expressions like $2^n+1$ or $10^n+1$ either. – lulu Sep 4 '17 at 11:09