a) Let Q(x) = x2−x+1. Show that if m is an integer greater than 1 such that m divides a, then m divides none of Q(a), Q(Q(a)), Q(Q(Q(a))), and so on. (b) Use part (a) to prove that there are infinitely many primes.

@mathmagician Try this :P

if a is even it is obvious, because then m is also must be even and Q(a) is not even, so m doesnt divide Q(a). Q(Q(a)) then wuld be also not even: odd number squared is odd, odd number minus odd number is even, and even number +1 is odd. So, a must be odd. Now, i'll try to analyze odd numbers, hope, will get the answer :)

As for odd numbers take any odd number 2k+1 and suppose, that m divides 2k+1. Then, when you put 2k+1 to Q(a), you get 2k(2k+1)+1. Since m divides 2k(2k+1), it cannot divide 2k(2k+1)+1. When you calculate Q(Q(a)), you get 12k^4+(2k+1)(8k^2+1). m divides (2k+1)(8k^2+1), but cannot divide (2k+1)(8k^2+1), since m is odd.

Besides, i have a feeling that Q(a) generates only prime numbers, but I cannot prove it. Besides, i've heard a theorem (or hypothesis) that any polynomial of second order can generate only prime numbers (despite the fact, that some polynomials generate the more often than the others).

Join our real-time social learning platform and learn together with your friends!