Let $A$ be a set satisfying the statement of the problem, with $|A| \geq 3$. Clearly, $1 \in A$ from a) and since $A$ is nonempty. - We have $2 \in A$ : Suppose that it is not the case. Then from $|A| \geq 3$, it follows that there are odd integers $a,b \in A$ such that $1<a<b$. From b), we deduce that $1+ab \in A$. Since it is an even number, we must have $2 \in A$. A contradiction. - Let $p \in A$ with $p >2$. First note that such a $p$ does exist since $|A| \geq 3$. Using b), we deduce that $1+2p \in A$. Thus $1 + 2(1+2p) = 4p+3 \in A$, and by an easy induction $2^np + 2^n - 1\in A$ for all integers $n \geq 0$. (1) Moreover, $1+p(1+2p) = 1 + p + 2p^2 \in A$ so that $1 + 2(1+p+2p^2) = 4p^2 + 2p + 3 \in A$. - Now we prove that $4 \in A$ : Let $p \in A$ with $p > 2$. If $p = 0$ mod[4], we are done from a). If $p = 2$ mod[4], then using a) and since $p > 2$, we may assume that $p$ is odd, by replacing ^p^by one of its odd divisors. If $p = 1$ mod[4], then $2p^2 + p + 1 = 0$ mod[4], and we are done again. If $p = 3$ mod[4], then $4p^2 + 2p + 3 = 9 = 1$ mod[4]. Thus we are left to the previous case, and we are done. - It follows that $3 \in A$ : Since $4 \in A$, it follows that $1+2 \times 4 = 9 \in A$. Thus $3 \in A$. - Now we prove that each odd poditive integer is in $A$ : From (1) and since $3 \in A$, we deduce that for each $n \geq 0$, we have $3 \times 2^n + 2^n - 1 = 2^{n+2} - 1 \in A$. (2) Let $a$ be a positive odd integer, with $a \geq 5$. Since $a$ is coprime with 2, it follows that there exist an integer $n \geq 0$ such that $2^{n+2} = 1$ modmath=inline]$a$[/math. Using (2) and a), we deduce that $a \in A$. - Let $a \geq 5$ be odd. Then from a), $1+3a \in A$. Thus $A$ contains each sufficiently large even integer equals to 1 mod[3]. Since $(3a+1)^2 = 1$ mod[3], it follows from a) that $A$ contains each integer equals to 1 mod[3]. Moreover, since $(3a+2)^{2n} = 1$ mod[3] for all integers $n$, it follows from a) that $A$ contains each integer equals to 2 mod[3]. Now let $a$ be an even integer equals to 0 mod[3]. Since $a^n - 1 = (a-1)(a^{n-1} + ... + 1)$, we may choose $n$ such that $a^n - 1$ is not a prime or the square of a prime. For such a $n$, there exists two odd integers $p,q$ such that $1<p<q$ and $p,q \in A$ from above. Thus $a^n = 1 + pq \in A$, so that from a), we have $a \in A$. Then $A$ contains all positive integers, and we are done. Pierre.