# Division of integers by 5 [closed]

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 $dn^3+cn^2+bn+a\equiv0\bmod5$.

• What did you try? May 26, 2017 at 12:06
• Hint: $$ax^3+bx^2+cx+d=x^3\left(d(\frac1x)^3+c(\frac1x)^2+b\frac1x+a\right).$$ May 26, 2017 at 12:11
• 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. May 26, 2017 at 16:53

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 $a_0 x^n + a_{1} x^{n-1} + \cdots + a_{n-1} x + a_n$.

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!

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

$$am^3+bm^2+cm+d\equiv0\bmod5$$ $m$ 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: $$dn^3+cn^2+bn+a\equiv0\bmod5$$

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 $n^3$.