Nice Number Theory

Warning: Gigantic solution ahead. Brace yourselves...

Okay, so, first of all, it should be clear that $n$ and $n + 2$ must have the same parity.

Case $I$ : $n$ is even.

If $n$ is even, then so is $n + 2$ ; thus, $n = 2p_{1}$ and $n + 2 = 2p_{2}$ , where $p_{1}$ and $p_{2}$ are prime numbers.

From this, we can write: $2p_{2} - 2p_{1} = 2$ , or $p_{2} - p_{1} = 1$ . This implies that both primes are consecutive, and thus $p_{1} = 2$ and $p_{2} = 3$ , which are the only possibilities; however, since $2$ is already a factor of $n$ , then this number is not the product of two distinct primes, so by contradiction n cannot be even.

Case $II$ : $n$ is odd.

If $n$ is odd, then so is $n + 2$ , and $n + 1$ is even. From this, we can write $n + 1 = 2p$ , where $p$ is an odd prime number.

Now, we use a result that every prime greater than 4 modulo $6$ is either $1$ or $5$ (The only numbers less than six which are coprime to it).

Thus, $2p \equiv 2 \pmod{6}$ or $2p \equiv 10 \equiv 4 \pmod{6}$ , so either $n \equiv 1 \pmod{6}$ or $n \equiv 3 \pmod{6}$

Now, if $n = p_{1}*p_{2}$ where $p_{1}$ and $p_{2}$ are primes, then we have two cases. WLOG, let's assume that $p_{1} < p_{2}$ ;

Subcase $1.1$ : $p_{1} = 3$

We have that either $p_{2} \equiv 1 \pmod{6}$ or $p_{2} \equiv 5 \pmod{6}$ ; thus, $n \equiv 3 \pmod{6}$ or $n \equiv 15 \equiv 3 \pmod{6}$ ; in either case, if one of the factors of $n$ is $3$ , then $n \equiv 3 \pmod{6}$ ; also, $n + 2 \equiv 5 \pmod{6}$

Subcase $1.2$ : $p_{1} \neq 3$

Then, by reasoning similar to the one above, $n \equiv 1 \pmod{6}$ or $n \equiv 5 \pmod{6}$ ; due to the conditions set above, we can only have $n \equiv 1 \pmod{6}$ ; this can only happen when both primes leave the same remainder when divided by 6; this also implies that $n + 2 \equiv 3 \pmod{6}$

This proves that at least one of $n$ and $n + 2$ is divisible by $3$ . (Both are odd numbers, and the set of numbers $\equiv 3 \pmod{6}$ is $(3, 9, 15, \ldots)$ )

From here on, it's pretty much trial and error. Firstly we'll work with the case $n = 3q$ , and then later with the case $n + 2 = 3r$ , both q and r being prime numbers.

Subcase $2.1$ : $n = 3q$

$q = 5 \rightarrow n = 15$ ; $n + 1 = 16 \neq 2t$ , where $t$ is a prime number.

$q = 7 \rightarrow n = 21$ ; $n + 1 = 22 = 2*11$ ; $n + 2 = 23 \neq p_{i}*p_{j}$ , which is not a valid solution.

$q = 11 \rightarrow n = 33$ ; $n + 1 = 34 = 2*17$ ; $n + 2 = 35 = 5*7$ ; $n = 33$ is a valid member of the set.

$q = 13 \rightarrow n = 39$ ; $n + 1 = 40 \neq 2*t$ , where $t$ is a prime number.

$q = 17 \rightarrow n = 51$ ; $n + 1 = 52 = 2^{2}*13 \neq 2*t$ , where $t$ is a prime number.

$q = 19 \rightarrow n = 57$ ; $n + 1 = 58 = 2*29$ ; $n + 2 = 59 \neq p_{i}*p_{j}$ , which is not a valid solution

$q = 23 \rightarrow n = 69$ ; $n + 1 = 70 = 2*5*7 \neq 2*t$ , where $t$ is a prime number.

$q = 29 \rightarrow n = 87$ ; $n + 1 = 88 = 2^{3}*11 \neq 2*t$ , where $t$ is a prime number.

$q = 31 \rightarrow n = 93$ ; $n + 1 = 94 = 2*47$ ; $n + 2 = 95 = 5*19$ ; $n = 93$ is a valid member of the set.

$q = 37 \rightarrow n = 111$ ; $n + 1 = 112 = 2^{4}*7 \neq 2*t$ , where $t$ is a prime number

$q = 41 \rightarrow n = 123$ ; $n + 1 = 124 = 2^{2}*31 \neq 2*t$ ,where $t$ is a prime number

$q = 43 \rightarrow n = 129$ ; $n + 1 = 130 = 2*5*13 \neq 2*t$ , where $t$ is a prime number

$q = 47 \rightarrow n = 141$ ; $n + 1 = 142 = 2*71$ ; $n + 2 = 143 = 11*13$ ; $n = 141$ is a valid member of the set.

$q = 53 \rightarrow n = 159$ ; $n + 1 = 160 = 2^{5}*5 \neq 2*t$ , where $t$ is a prime number

$q = 59 \rightarrow n = 177$ ; $n + 1 = 178 = 2*89$ ; $n + 2 = 179$ is a prime number

$q = 61 \rightarrow n = 183$ ; $n + 1 = 184 = 2^{3}*23 \neq 2*t$ , where $t$ is a prime number

$q = 67 \rightarrow n = 201$ ; $n + 1 = 202 = 2*101$ ; $n + 2 = 203 = 7*29$ ; $n = 201$ is a valid member of the set

And so on...

Subcase $2.2$ : $n + 2 = 3r$

$q = 5 \rightarrow n = 13$ ; $n$ is not the product of two distinct prime numbers

$q = 7 \rightarrow n = 19$ ; $n$ is not the product of two distinct prime numbers

$q = 11 \rightarrow n = 31$ ; $n$ is not the product of two distinct prime numbers

$q = 13 \rightarrow n = 37$ ; $n$ is not the product of two distinct prime numbers

$q = 17 \rightarrow n = 49 = 7^{2}$ ; $n$ is not the product of two distinct prime numbers

$q = 19 \rightarrow n = 55 = 5*11$ ; $n + 1 = 56 = 2^{3}*7 \neq 2*t$ , where $t$ is a prime number

$q = 23 \rightarrow n = 67$ ; $n$ is not the product of two distinct prime numbers

$q = 29 \rightarrow n = 85 = 5*17$ ; $n + 1 = 86 = 2*43$ ; $n + 2 = 87 = 3*29$ ; $n = 85$ is a valid member of the set.

$q = 31 \rightarrow n = 91 = 7*13$ ; $n + 1 = 92 = 2^{2}*23 \neq 2*t$ , where $t$ is a prime number

$q = 37 \rightarrow n = 109$ ; $n + 1 = 110 = 2*5*11 \neq 2*t$ , where $t$ is a prime number

$q = 41 \rightarrow n = 121 = 11^{2}$ ; $n$ is not the product of two distinct prime numbers

$q = 43 \rightarrow n = 127$ ; $n$ is not the product of two distinct prime numbers

$q = 47 \rightarrow n = 139$ ; $n$ is not the product of two distinct prime numbers

$q = 53 \rightarrow n = 157$ ; $n$ is a prime number.

$q = 59 \rightarrow n = 175 = 5^{2}*7$ ; $n$ is not the product of two distinct prime numbers

$q = 61 \rightarrow n = 181$ ; $n$ is a prime number

$q = 67 \rightarrow n = 199$ ; $n$ is a prime number

And so on...

The first members of the set $S$ are: $(33, 85, 93, 141, 201,\ldots)$ ; thus, the product sought is $33*85*93*141*201 = 7393174965$

The numbers in question are :

$n= 33 (n=3\times11 ;n+1=2\times17;n+2=5\times7)$

$n= 85 (n=5\times17 ;n+1=2\times43;n+2=3\times29)$

$n= 93 (n=3\times31 ;n+1=2\times47;n+2=5\times19)$

$n= 141 (n=3\times47 ;n+1=2\times71;n+2=11\times13)$

$n= 201 (n=3\times67 ;n+1=2.101;n+2=7\times29)$ .

I found this question in a book where solutions were not given, I'd love if someone else posts their solution !!!!