If n is positive integer then 2n^4 - 41n^3 + 18n^2 + 63n + 23, 40n^4 - 89n^3 + 64n^2 + 99n + 57 are not both prime. Proof or counterexample?

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.

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