<s>[INST] Prove the polynomial $f(x) = 3x^3+x^2+2x+1155$ has no root in $\mathbb{Z}$ Prove the polynomial $f(x) = 3x^3+x^2+2x+1155$ has no root in $\mathbb{Z}$
The hint says If $f$ has a root in $\mathbb{Z}$, then $f$ has also a root in $\mathbb{Z}/2\mathbb{Z}$. I'm still confused  
 [/INST] If $f(x) = 0$ then $f(x) \equiv 0 \mod n$ for all $n$.
So $f(x) \equiv 0 \mod 2$.
So $3x^3+x^2+2x+1155 \equiv 0 \mod 2$
$x^3 + x^2 + 1  \equiv 0 \mod 2$
$x^3 + x^2 \equiv 1 \mod 2$
Notice $0^k \equiv 0 \mod n$ and $1^k \equiv 1 \mod n$ and and $x$ is either $x \equiv 0 \mod 2$ or $x \equiv 1 \mod2$ we have $x^k \equiv x \mod 2$.
So 
$x^3 + x^2 \equiv x+x = 2x \equiv 0 \mod 2$ 
This contradicts $x^3 + x^2 \equiv 1 \mod2$.
....
Actually if we had started with the observation $x^k \equiv x \mod 2$ we could have done simply
$3x^3 + x^2 +2x + 1155 \equiv$
$3x + x +2x +1155 \equiv$
$6x + 1155 \equiv$
$1155 \equiv$
$1 \not \equiv 0 \mod 2$.
....
Alternatively if we never made the observation $x^k \equiv x \mod 2$ we could have broken 
$x^3 + x^2 \equiv 1 \mod 2$ into two cases: $x \equiv 1 \mod 2$ and $x\equiv 0 \mod 2$ and not Case 1 yields $1 + 1 \equiv 0 \mod 2$ and the second $0+0 \equiv 0 \mod 2$.
Or we could have noted:
$x^3 + x^2 = x^2(x+1)$ and note them most be of different parity.
.....
Or if on this $\mod 2$ notation doesn't same natural, we could have simply said from the start.
If $x \in \mathbb Z$ either $x$ is even or odd.
If $x$ is even $mx^k$ is even so 
$3x^2 + x^2 + 2x + 1115 = even + even + even + odd = odd \ne 0$.
If $x$ is odd then $mx^k$ is odd if $m$ is odd and $mx^k$ is even if $m$ is even so 
$3x^2 + x^2 + 2x + 1115 = odd + odd + even + odd = odd \ne 0$.
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And if you want to get really annoying...
Let $x = 2y + i$ where $i$ is either $1$ or $0$.
$f(x) = 3(2y+i)^3 + (2y + i)^2 + 2(2y+i) + (2k+1)$ where $k = \frac {1155 -1}2$.
$= EVEN + 3i^3 + EVEN + i^2 + EVEN  + 1$
$= EVEN + (3i^3 + i^2) + 1$
$= EVEN + (\{0 + 0|3 + 1\}) + 1$
$= EVEN + 1$
$= ODD \ne 0$.
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