1) is Obvious the kobayashy theorem 2)$p=4k+3$ and $a$ is not a quadratic residue $(mod p)$ and use the Chinese Remainder Theorem.

At first sight it seems you're quite cheating Point i), is it really necessary to quote the Kobayashi theorem? (A sketch of proof here) Point ii) what about the case $a=4$? Ps. Dan, point i) is a corollary if this one, do you remember it?

You can obviously replace Kobayashi by Zsigmondy Or for simplicity set $n$ to be a large prime, then any prime divisor $q$ of the number will be $\equiv 1 \mod p$ or dividing $p(a+1)$. Now either use several primes $n$ at once or do some work $\mod q^2$ to show that this will give you new ones.

I think that this can also be done quite an easy way. For i) Assume that ${a^n+1}$ has finite prime divisors: $q_1,...,q_k$. Then $a^n+1=q_1^{a_1}...q_k^{a_k}$, denote $A_n$ the largest among ${a_1,...,a_k}$. As n increases, it can be seen that $A_n$ must increase boundlessly. For any k+1 successive numbers of the form $a^i+1$, there must be 2 such that its maximum value $A$ is the exponent of the same prime (among $q_1,...,q_k$). Thus, we get $a^t+1-(a^j+1) \vdots (q_l)^A$. As $(a,q_l)=1$, $a^{t-j}-1 \vdots (q_l)^A$. However, ${t-j}={1;2;...;k}$ yet $(q_l)^A$ approaches infinity, we arrive at contradiction. ii) Take $h_i$ as a prime divisor of $a^{2i+1}-1$. Then $ord_a(h_i)$ is odd and thus, there is no n such that $a^n+1 \vdots h_i$. But similar to the above proof, the sequence $a^{2i+1}-1$ must also have infinite prime divisors.

My solution: a)It's trivial that by zigmondy's theorem $a^n+1$ has a prime factor that it doesn't divide $a^i+1$ for $1\le i\le n-1$. b)Take an arbitrary odd number $n$ and a prime $p$ such that $p\mid a^n-1$ we show that there isn't any $m\in N$ such that $p\mid a^m+1$ assume for a sake of a contradiction that $p\mid a^m+1$ then $a^{2m}\equiv 1\pmod{p}$(1) let $d$ order of $a$ modulo $p$ then from (1) we get that $d$ must be even but because $a^n\equiv 1\pmod{p}\longrightarrow d\mid n$ but it's a contradiction ($d$ is even and $n$ is odd) from zigmondy's theorem the number of these primes are infinite.

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