Show that for any prime $p$ , there are either infinity many or no positive integer $a$ , such that 6p divides $a^p +1$. Find all those primes for which there exists no solution.
Answer:
Suppose, $\space a = 6pn-1 , where \space n\in N \\ \implies a \space \equiv -1 \space(mod\space6p)\\\implies a^p\equiv (-1)^p (mod\space6p)$ $ \\if \space p\neq2,\space a^p\equiv -1(mod\space 6p)\\ \space \space \space \implies \space a^p+1\equiv0(mod\space 6p) \\so,\space6p \vert \space a^p +1$
As there are infinitely many values of $n$ for which $a^p+1$ Divides 6p , we can say that For all primes other than 2 there exists infinitely many solutions.
Now if, $p=2, \\ a^p+1=a^2 +1 \space and \space 6p=12$
But for any positive integer $a$, $\\a^2 \equiv 0\space (mod\space 4)\space or\space a^2\equiv 1 (mod\space 4) \\ \implies a^2+1\equiv 1 (mod4) \space \\\space or \space a^2+1 \equiv 2(mod 4) \\So, a^2+1 \space cannot\space divide \space 12\space as \space it\space \\doesn't \space divide \space 4 .$
No solution exists for $a$ if $p=2$ $$QED$$