SOLUTION: Mathematicians have been searching for a formula that yields prime numbers. One such formula was x^2-x+41. Select some numbers for x, substitute them in the formula, and see if pr

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Mathematicians have been searching for a formula that yields prime numbers. One such formula was x^2-x+41. Select some numbers for x, substitute them in the formula, and see if pr      Log On


   



Question 557458: Mathematicians have been searching for a formula that yields prime numbers. One such formula was x^2-x+41. Select some numbers for x, substitute them in the formula, and see if prime numbers occur. Try to find a number for x that when substituted in the formula yields a composite number.
I'm not sure of the steps to complete this problem...Please explain.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The polynomial P%28x%29=x%5E2-x%2B41 was supposed to yield prime numbers.
You are expected to try a few values for x , and find the corresponding P(x).
It is likely to be a prime number. Here are a few:
P(0)=41, P(1)=41, P(2)=43, P(3)=47, P(4)=53, P(5)=61, P(6)=71, P(7)=83, P(8)=97, P(9)=113, P(10)=131 P(20)=421, P(30)=971, P(40)=1601.
All of those (and P%28x%29 for the x values in between) are prime numbers.
It was a nice try, the design of P(x) ensured that iy could not be a multiple of 2, 3, 5, 7.
However,
P%28x%29=x%5E2-x%2B41=x%28x-1%29%2B41, so if x or x-1 were a multiple of 41, P%28x%29 would be a multiple of 41. So,
P%2841%29=41%2A40%2B41=41%2A%2840%2B1%29=41%2A41
P%2842%29=42%2A41%2B41=41%2A%2842%2B1%29=41%2A43
P%2882%29=82%2A81%2B41=2%2A41%2A81%2B41=41%2A%282%2A81%2B1%29=41%2A163
P%2883%29=82%2A83%2B41=2%2A41%2A83%2B41=41%2A%282%2A83%2B1%29=41%2A167, and so on
There are other values of P(x) that are not prime, too, like
P%2845%29=2021=43%2A47
P%2850%29=2491=81%2A53
P%2866%29=4331=61%2A71
P%2877%29=5893=71%2A83
P%2885%29=7181=43%2A167