wrote:>

If a,b,c are not required to be primes, then there are many solutions.

> Let f(x)=x^2-x+1.

> Find 2 triplets, {a,b,c}, of primes

> such that abc|f(a)f(b)f(c).

>

If a<b<c, the one with minimum a is:

a=31 => f(a)=931=7^2*19

b=133=7*19 => f(b)=17557=97*181

c=181 => f(c)=32581=31*1051

-Mike Oakes