San
Favorite Answer
2n^4 - 41n^3 + 18n^2 + 63n + 23, 40n^4 - 89n^3 + 64n^2 + 99n + 57 are not both prime.
Because either 2n^4 - 41n^3 + 18n^2 + 63n + 23 or 40n^4 - 89n^3 + 64n^2 + 99n + 57 has 3 as a factor.
Proof:
2n^4 - 41n^3 + 18n^2 + 63n + 23
= 3n^4 -42n^3 +18n^2 +63n +24 -n^4 + n^3 -1
= 3(n^4 -14n^3 +6n^2 +21n +8) + (-n^4 + n^3 -1)
40n^4 - 89n^3 + 64n^2 + 99n + 57
= 39n^4 -90n^3 +63n^2 +99n +57 + n^4 +n^3 +n^2
= 3(13n^4 -30n^3 +21n^2+33n +19) + n^2(n^2 +n +1)
Therefore, if -n^4 + n^3 -1 has 3 as a factor, 2n^4 - 41n^3 + 18n^2 + 63n + 23 has 3 as a factor as well.
In the same way, if n^2(n^2 +n +1) has 3 as a factor, 40n^4 - 89n^3 + 64n^2 + 99n + 57 has 3 as a factor, too.
Consider these three patterns: n=3m, 3m-1, 3m-2 (m is a positive integer).
n=3m
n^2(n^2 +n +1)
=9m^2(9m^2 +3m +1)
=3(27m^4 +9m^3 +3m^2)
As n^2(n^2 +n +1) has 3 as a factor, 40n^4 - 89n^3 + 64n^2 + 99n + 57 has 3 as a factor. 40n^4 - 89n^3 + 64n^2 + 99n + 57 isn't a prime number.
n=3m-1
-n^4 + n^3 -1
=-(3m-1)^4 +(3m-1)^3 -1
=-81m^4 +135m^3 -81m^2 +27m -3
=3(-27m^4 + 45m^3 -27m^2 +9m -1)
As -n^4 + n^3 -1 has 3 as a factor, 2n^4 - 41n^3 + 18n^2 + 63n + 23 has 3 as a factor. 2n^4 - 41n^3 + 18n^2 + 63n + 23 isn't a prime number.
n=3m-2
n^2(n^2 +n +1)
=(3m-2)^2[(3m-2)^2 +(3m-2) +1]
=(3m-2)^2 (9m^2 -9m +3)
=3(3m-2)^2 (3m^2 -3m +1)
As n^2(n^2 +n +1) has 3 as a factor, 40n^4 - 89n^3 + 64n^2 + 99n + 57 has 3 as a factor. 40n^4 - 89n^3 + 64n^2 + 99n + 57 isn't a prime number.
In all cases, either 2n^4 - 41n^3 + 18n^2 + 63n + 23 or 40n^4 - 89n^3 + 64n^2 + 99n + 57 has 3 as a factor. Therefore either of them isn't a prime number.
Jim
n=1
2-41+18+63+23 = 65 which is not prime
therefore false
Again trying n=1:
40 -89 +64 +00 +57 = 72
Again this is not prime