Indija TS 2012

If $S$ is singleton, then the statement is vacuously true. Hence $S=\{p\}$, $p$ a prime.

Suppose $S$ contains two elements. One of these must be odd primes and hence $2$ is in $S$. Moreover $p-1$ must be a power of $2$. Hence $p$ is Fermat prime.

Suppose $|S|\ge 3$. Here we consider two cases: $|S|<\infty$ and $S$ is an infinite set.

If $|S|<\infty$, let $S=\{p_1(=2),p_2,\dots ,p_r\}$, with $r\ge 3$. Since $|S|\ge 3$, there is a prime of the form $3k+1$ or $3k+2$ in $S$. Observe $(3k+1)-1=3k$ and $2(3k+2)-1=6k+3$ both have $3$ as a prime factor. Hence $3\in S$. Now $$(2\cdot p_3\cdots p_r)-1=3^k\text{ and }(3p_3\cdots p_r)-1=2^l,$$ for some $k>0$ and $l>0$. Hence we get $$3(3^k+1)=2(2^l+1).$$ This gives $2^{l+1}-3^{k+1}=1$. The only possibility is $l+1=2$ and $k+1=1$. This forces $k=0$, a contradiction. Hence $3\le |S|<\infty$ is not possible.

Finally assume $S$ is an infinite set. Suppose there exists a prime $p$ not in $S$. Take $S=\{p_1(=2),p_2,\dots \}$

Consider $p_1-1,p_1{p}_2-1,\dots ,p_1{p}_2\dots p_p-1$ modulo $p$. Since $p$ is not in $S$, none of them is equal to $0$. By pigeonhole principle, we can find $j<k$ such that $$p_1{p}_2\cdots p_j-1\equiv p_1{p}_2\cdots p_j{p}_{j+1}\cdots p_k-1(\mathrm{mod}p).$$ Hence $p$ divides $p_1{p}_2\cdots p_j(p_{j+1}\cdots p_k-1)$. But this contradicts the choice of $p$ that it is not in $S$. Hence $S$ is the set of all primes now.