Quadratic primes Euler discovered the remarkable quadratic formula: n ² + n + 41 It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 40 2 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41. The incredible formula n ² − 79 n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, −79 and 1601, is −126479. Considering quadratics of the form: n ² + an + b , where | a | < 1000 and | b | < 1000 where | n | is the modulus/absolute value of n e.g. |11| = 11 and |−4| = 4 Find the product of the coefficients, a and b , for the quadratic expression that produces the maximum number of primes for consecutive values of n , starting with n = 0. Before I start solving this problem, I would like to apologize for not posting the solution for these many days. I will tr...
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