## Quadratic primes

### Problem 27

Published on Friday, 27th September 2002, 06:00 pm; Solved by 49735Euler 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:

n² +an+b, where |a| < 1000 and |b| < 1000

where |n| is the modulus/absolute value ofn

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.