2007-03-03 14:50:52 · 7 answers · asked by Faraz S 3 in Science & Mathematics ➔ Mathematics
Well, that's false, since 41 is prime but obviously 41^2 + 41 + 41 is not prime. That factors to 41 * (41 + 1 + 1) = 41 * 43
HTH
Charles
2007-03-03 14:56:49 · answer #1 · answered by Charles 6 · 4⤊ 0⤋
All values of n that are a positive multiple of 41 will result in a value that is non-prime that is divisible by 41.
n = 1 * 41 = 41
(41)^2 + 41 + 41 = 1763
→ 1763/41 = 43
n = 2 * 41 = 82
(82)^2 + 82 + 41 = 6874
→ 6874/41 = 167
n = 3 * 41 = 123
(123)^2 + 123 + 41 = 15293
→ 15293/41 = 373
...
2007-03-05 09:20:37 · answer #2 · answered by Kookiemon 6 · 0⤊ 0⤋
n=1
1^2+1+41=43 is a prime number.
n=2
2^2+2+41=47 is a prime number
n=3
3^3+3+41=53 is a prime numbrer
and so on.
for any value of n we get prime number.
2007-03-03 17:55:30 · answer #3 · answered by shohag 1 · 0⤊ 2⤋
40^2 + 40 + 41 = 1681 = 41^2, which is not a prime number.
2007-03-03 15:00:20 · answer #4 · answered by MHW 5 · 4⤊ 0⤋
It is disproved by trial and error, the errors occurring at n = 40 and n = 41.
The predominance of primes up to the first error is related to the peculiar properties of (4*41 - 1) = 163. For example,
exp (pi * sqrt(163)) = 262 53741 26407 68743.99999 99999 9925...
which is remarkably close to a whole number.
2007-03-04 03:44:38 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Hello,
The statement is false.
Let n=41.
Then 41^2 + 41 + 41 is divisible by 41.
I got burned by this question on an exam.
2007-03-03 15:00:46 · answer #6 · answered by toyallhi 2 · 3⤊ 0⤋
n^2 +n+1 also equals a prime number
2007-03-03 15:01:06 · answer #7 · answered by Anonymous · 0⤊ 5⤋