Can someone explain me why $\displaystyle n^2+n+1$ isn't dividable with 125?
Consider $\displaystyle n^2+n+1$ modulo $\displaystyle 10$, for $\displaystyle n=10k+r$ withOriginally Posted by DenMac21
$\displaystyle r\in \{0,1,2,3,4,5,6,7,8,9\}$. You will find $\displaystyle n^2+n+1$
is never divisible by $\displaystyle 5$ and so cannot be divisible by $\displaystyle 125$.
RonL
If,Originally Posted by DenMac21
$\displaystyle n^2+n+1\equiv 0\mod 5$
Noticing that 5 is a prime, look at its discrimanant ($\displaystyle b^2-4ac)$.
The above congruence has a solution if and only if its discrimant is a quadradic residue.
Since $\displaystyle b^2-4ac=-3\equiv 2$.
Apply Euler's Criterion to determine if something is a quadradic residue we get that,
$\displaystyle 2^{\frac{5-1}{2}}\not \equiv 1\mod 5$
Thus, $\displaystyle n^2+n+1$ is never divisible by 5, thus how can it be divisible by $\displaystyle 125=5^3$?
Thus, we have shown it is impossible.
Q.E.D.
To remind you Euler's Criterion states, $\displaystyle a$ is a quadradic residue of $\displaystyle p$ if and only if,
$\displaystyle a^{\frac{p-1}{2}}\equiv 1\mod p$