Question 557458
The polynomial {{{P(x)=x^2-x+41}}} 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(x)}}} 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(x)=x^2-x+41=x(x-1)+41}}}, so if {{{x}}} or {{{x-1}}} were a multiple of 41, {{{P(x)}}} would be a multiple of 41. So,
{{{P(41)=41*40+41=41*(40+1)=41*41}}}
{{{P(42)=42*41+41=41*(42+1)=41*43}}}
{{{P(82)=82*81+41=2*41*81+41=41*(2*81+1)=41*163}}}
{{{P(83)=82*83+41=2*41*83+41=41*(2*83+1)=41*167}}}, and so on
There are other values of P(x) that are not prime, too, like
{{{P(45)=2021=43*47}}}
{{{P(50)=2491=81*53}}}
{{{P(66)=4331=61*71}}}
{{{P(77)=5893=71*83}}}
{{{P(85)=7181=43*167}}}