Problem 27

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:

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.

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These problems are part of
Project Euler
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