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Problem 27


Problem 27


Quadratic primes

Euler discovered the remarkable quadratic formula:

n2 + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 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 n2 − 79n + 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:

  • n2 + 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.


二次“素数生成”多项式

欧拉发现了这个著名的二次多项式:

n2 + n + 41

对于连续的整数n从0到39,这个二次多项式生成了40个素数。然而,当n = 40时,402 + 40 + 41 = 40(40 + 1) + 41能够被41整除,同时显然当n = 41时,412 + 41 + 41也能被41整除。

随后,另一个神奇的多项式n2 − 79n + 1601被发现了,对于连续的整数n从0到79,它生成了80个素数。这个多项式的系数-79和1601的乘积为-126479。

考虑以下形式的二次多项式:

  • n2 + an + b, 满足|a| < 1000且|b| < 1000

  • 其中|n|指n的模或绝对值
    例如|11| = 11以及|−4| = 4

这其中存在某个二次多项式能够对从0开始尽可能多的连续整数n都生成素数,求其系数a和b的乘积。