Prime factors of $5 n^4 - 70 n^3 + 380 n^2 - 945 n + 911 $

119 Views Asked by At

Let $n$ be an integer. Then any prime factor of

$$ 5 n^4 - 70 n^3 + 380 n^2 - 945 n + 911 $$

Must be congruent to 1 mod 10.

Also

Let $n$ be an integer. Then any prime factor of

$$ 5 n^4 - 10 n^3 + 20 n^2 - 15 n + 11 $$

Must be congruent to 1 mod 10.

How to prove these ?

How to find such identities ?

1

There are 1 best solutions below

7
On

Hint: By factoring the first polynomial minus one we get, by setting $p(n)=5 n^4 - 70 n^3 + 380 n^2 - 945 n + 911$,

$$ 4\cdot p(n) = -1+5\left(2n^2-14n+27\right)^2 $$ so for every odd prime divisor $q\mid p(n)$ we have $\left(\frac{5}{q}\right)=+1$ and by quadratic reciprocity $q$ is forced to belong to some particular residue classes $\!\!\pmod{20}$. It is not difficult to show that $p(n)$ is odd for any $n\in\mathbb{Z}$ and similar arguments apply to the second polynomial, too.