Check your Number Theory Skills

If for a positive integer n the number $n^{2}-1$ is a product of three different primes, then (in view of $2^{2} -1 = 3$ ) we have n > 2. Next, in view of the identity $n^{2}-1 = (n-1)(n+1)$ , the number n must be even since otherwise the factors on the right-hand side would be even, and $2^{2}|n^{2}-1$ .

Moreover, the numbers n -1 and n + 1 (which are both > 1 since n > 2) cannot be both composite since in this case $n^{2}-1$ could not be a product of three different primes. Thus, one of the numbers n -1 and n + 1 must be a prime, and the other one must be a product of two primes.

For n = 4, we get n -1 = 3, n + 1 = 5, and this condition is not satisfied. Similarly, for n = 6, we get n -1 = 5, n +1 = 7; for n = 8, we have n -1 = 7, n +1 = 9 = $3^{2}$ . For n = 10, we have n -1 = 32, and for n = 12 we have n -1 = 11, n +1 = 13.

For n = 14, we have n -1 = 13, n +1 = 15 = 3· 5. Thus, the least positive integer n for which $n^{2}-1$ is a product of three different primes is n = 14, for which $n^{2}-1$ = 3 · 5 · 13. Since $16^{2}-1$ = 3 · 5 ·17, we see that the next number which satisfies the required property is n = 16. Now, $18^{2}-1$ = 17 .19, $20^{2}-1$ = 19 .21 = 3· 7 ·19, and the third such number is n = 20. Next, $22^{2}-1$ = 3 · 7· 23, and the fourth of the required numbers is n = 22. Continuing in this way we find easily the fifth such number to be n = 32, for which $32^{2} -1$ = 3 · 11 · 31.

Thus, the first five integers n for which $n^{2}-1$ is a product of three different primes are 14, 16, 20, 22, and 32.

Thus $14 \times 16 \times 20 \times 22 \times 32$ is 3153920.

factorize so you get $(n+1)(n-1)$ , then you can brute force :v