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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?

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    $\begingroup$ $a+b+c$ 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. $\endgroup$ – Mark Bennet Apr 5 at 21:30
  • $\begingroup$ @MarkBennet Oh yes, i missed that, but in that case polynomial is still odd for all cases so proof would still hold $\endgroup$ – 1qwertyyyy Apr 5 at 21:33
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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 $x$.

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Hint: For polynomials with integer coefficients, $ a- b \mid P(a) - P(b)$.

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