-1
$\begingroup$

Given integers $a,b,c,d$ with $d\not\equiv0\bmod5$ and $m$ an integer for which $am^3+b m^2+cm+d\equiv0\bmod5$, prove that there exists an integer $n$ for which $dn^3+cn^2+bn+a\equiv0\bmod5$.

$\endgroup$
3
  • $\begingroup$ What did you try? $\endgroup$ May 26, 2017 at 12:06
  • $\begingroup$ Hint: $$ax^3+bx^2+cx+d=x^3\left(d(\frac1x)^3+c(\frac1x)^2+b\frac1x+a\right).$$ $\endgroup$ May 26, 2017 at 12:11
  • $\begingroup$ To bring to the fore the innate symmetry in @Jyrki's comment, we can phrase it as follows: $r\neq 0$ is a root of a a polynomial $f(x)\iff$ its reciprocal $r^{-1}$ is a root if the reciprocal (reversed) polynomial. This symmetry has many nice applications, e.g. here. $\endgroup$ May 26, 2017 at 16:53

4 Answers 4

1
$\begingroup$

Hint: Consider the analogous statement over the reals:

If $r \in \mathbb R$ is a root of a real polynomial $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ with $a_0\ne0$, then $r\ne0$ and $1/r$ is a root of $a_0 x^n + a_{1} x^{n-1} + \cdots + a_{n-1} x + a_n$.

$\endgroup$
1
$\begingroup$

Since $5$ is prime, there exists an multiplicative inverse $n$, for $m$ such that $mn\equiv1\bmod5$
$$am^3+b m^2+cm+d\equiv0\bmod5$$ Multiply both sides by $n^3$ $$am^3n^3+b m^2n^3+cmn^3+dn^3\equiv0\bmod5$$ $$a+bn+cn^2+dn^3\equiv0\bmod5$$
Hence proved!

$\endgroup$
3
  • $\begingroup$ Why is $m \not \equiv 0 \bmod 5$? $\endgroup$
    – lhf
    May 26, 2017 at 12:14
  • $\begingroup$ @lhf otherwise that means $d\equiv0\bmod5$. OP states that this isn't true. $\endgroup$
    – maverick
    May 26, 2017 at 12:16
  • $\begingroup$ I know. I meant to suggest that you explain this in your answer... $\endgroup$
    – lhf
    May 26, 2017 at 12:38
1
$\begingroup$

$$am^3+bm^2+cm+d\equiv0\bmod5$$ $m$ is not zero because $d$ would then be zero. The modulus of 5 is prime, so $m$ has an inverse $n$: $$an^{-3}+bn^{-2}+cn^{-1}+d\equiv0\bmod5$$ Multiplying by $n^3$ yields the desired result: $$dn^3+cn^2+bn+a\equiv0\bmod5$$

$\endgroup$
0
1
$\begingroup$

Hint: Since $d\ne 0$, we have $m\ne 0$. Thus $m$ is invertible modulo 5. Take the inverse $n$ of $m$ modulo 5 and multiply the equation with $n^3$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .