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Prime Numbers: (2^n - 1) and (2^n + 1)
the factorizations come fromm doing polynomial long division of x^n-1 and x^n+1 with x^p-1 and x^p+1 if n=pq x^n-1 = (x^p-1)(x^{p(q-1)}+x^{p(q-2)}+\cdots +x^p+1) The other equation should read: x^n+1 = (x^p+1)(x^{p(q-1)}-x^{p(q-2)}+\cdots -x^p+1 ) With alternating signs...- Eigenstates
- Post #6
- Forum: Calculus and Beyond Homework Help