Prove the polynomial $f(x) = 3x^3+x^2+2x+1155$ has no root in $\mathbb{Z}$

Question

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

Answer 1

Apply the following

Parity Root Test $\ $ A polynomial $\,f(x)\,$ with integer coefficients has no integer roots when its $\rm\,\color{#0a0}{constant\,\ coefficient}\,$ and $\,\rm\color{#c00}{coefficient\,\ sum}\,$ are both odd.

Proof $\ $ If so then $\ \color{#0a0}{f(0)} \equiv 1\equiv \color{#c00}{f(1)}\,\pmod{\! 2},\ $ i.e. $\:f\:$ has no roots in $\,\Bbb Z/2 = $ integers mod $\,2,\,$ therefore $\,f\,$ has no integer roots (since if $\,f(n) = 0\,$ then $\,f(n\bmod 2)\equiv 0\pmod{\!2}\ $ by the Polynomial Congruence Rule)

Answer 2

Note that for all integers $x$, we have $$x^3 \equiv x \pmod {2}, x^2 \equiv x \pmod {2}$$ Thus $$3x^3+x^2+2x+1155 \equiv 6x+1155=2(3x+577)+1\equiv 1 \pmod {2}$$ So $f(x)$ is always odd for integers $x$. However, $0$ is even.So there exists no integer $x$ such that $f(x)=0$.

Answer 3

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 $k = \frac {1155 -1}2$.

$= EVEN + 3i^3 + EVEN + i^2 + EVEN + 1$

$= EVEN + (3i^3 + i^2) + 1$

$= EVEN + (\{0 + 0|3 + 1\}) + 1$

$= EVEN + 1$

$= ODD \ne 0$.