# Show that $p(x) = ax^3+bx^2+cx+d$ has no integer roots, where $a,b,c,d\in\mathbb{Z}$ and $p(0), p(1)$ are odd

Coefficients of polynomial $$P(x)=ax^3+bx^2+cx+d$$ are integers. Numbers $$P(0)$$ and $$P(1)$$ are odd. Show polynomial $$P(x)$$ has no roots that are integers.

My proof:

$$P(0)=d$$

$$P(1)=a+b+c+d$$ is odd then $$a+b+c$$ is even that means that two of numbers must be odd and one- even.

Let's investigate parity of that polynomial due to the parity of argument.

Let $$\alpha$$ be even integer.

$$a\alpha^3+b\alpha^2+c\alpha=\alpha(a\alpha^2+b\alpha+c)$$ this part is even, but adding $$d$$ makes $$P(\alpha)$$ an odd number.

Let $$\beta$$ be odd integer.

In this case $$a\beta^3+b\beta^2+c\beta$$ is also even, because even number times odd number is even, so we have two even numbers plus one odd. That means $$P(\beta)$$ is also odd.

$$\forall{x\in Z}:2\nmid P(x)$$. That proves my thesis since $$0$$ is even.

Could you show me other methods doing this proof? Is this proof correct?

• $a+b+c$ even can mean each is individually even. The parity argument is a good one though if you can get it to work. Try to look for the most efficient way of writing the proof. – Mark Bennet Apr 5 at 21:30
• @MarkBennet Oh yes, i missed that, but in that case polynomial is still odd for all cases so proof would still hold – 1qwertyyyy Apr 5 at 21:33

More succinctly: if $$x$$ and $$y$$ are integers, $$m$$ any positive integer, and $$P$$ is a polynomial with integer coefficients, $$P(x) \equiv P(y) \mod m$$ if $$x \equiv y \mod m$$. In particular if $$P(0) \equiv P(1) \equiv 1 \mod 2$$, then $$P(x) \equiv 1 \mod 2$$ for all integers $$x$$.
Hint: For polynomials with integer coefficients, $$a- b \mid P(a) - P(b)$$.