Post #2141607

This is where a new variant of the criterion can be applied, this time with the prime number 3. Indeed, modulo 3, one has f(T)=T^4+2T^2+1=(T^2+1)^2. So we are almost as in the initial criterion, but the polynomial T is not T^2+1. The first thing that makes this criterion apply is that T^2+1 is irreducible modulo 3. In this case, this is because -1 is not a square mod 3. The criterion also requires of variant of the condition on the derivative — it holds because the polynomial is not zero modulo (T^2+1, 9). Here, one has T^4-10T^2+1=(T^2+1)^2-12T^2 = (T^2+1)^2-12(T^2+1)+12 is equal to 3 modulo (T^2+1, 9). And so we have an Eisenstein-type proof that the polynomial T^4-10T^2+1 is irreducible over the integers. CQFD. #math