The discussion centers on determining the number of polynomials of degree 5 with distinct coefficients from the set {1, 2, 3, 4, 5, 6, 7, 8} that are divisible by the polynomial x² - x + 1. The final consensus indicates that there are 300 valid polynomials meeting these criteria. The solution involves analyzing the conditions derived from polynomial division and counting distinct sextuplets that satisfy specific equations. The calculations and reasoning were verified through both theoretical analysis and computational methods.
PREREQUISITESMathematicians, computer scientists, and students interested in polynomial algebra, combinatorial mathematics, and computational problem-solving will benefit from this discussion.
chisigma said:If 'polynomials with distinct coefficients' means 'different polynomials' the solution is easy: the requested number is equal to the number of polynomials of degree 3 that can be constructed with a set of 8 coefficients, i.e. $\displaystyle n= 8^{4}= 4096$...
Kind regards
$\chi$ $\sigma$
Start by dividing the fifth degree polynomial $ax^5+bx^4+cx^3+dx^2+ex+f$ by $x^2-x+1$: $$ax^5+bx^4+cx^3+dx^2+ex+f = (x^2-x+1)\bigl(ax^3 + (a+b)x^2 + (b+c)x + (d+c-a)\bigr) + (d+e-b-a)x + (a+f-c-d).$$ The conditions for the remainder to be zero are $$\boxed{\begin{array}{c}a+b = d+e, \\ a+f = c+d. \end{array}}$$ So you are looking for the number of sextuplets $(a,b,c,d,e,f)$ consisting of six distinct elements from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that satisfy the boxed conditions.jacks said:Find the number of polynomials of degree $5$ with distinct coefficients from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that are divisible by $x^2 - x + 1$
jacks said:Find the number of polynomials of degree $5$ with distinct coefficients from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that are divisible by $x^2 - x + 1$
chisigma said:Let's write the 5 degree polynomial as $\displaystyle P(x)= a_{5}\ x^{5} + a_{4}\ x^{4} + a_{3}\ x^{3} + a_{2}\ x^{2} + a_{1}\ x + a_{0}$. If $\displaystyle x^{2} - x +1$ divides P(x), then $\displaystyle e^{i\ \frac{\pi}{3}}$ as well as $\displaystyle e^{- i\ \frac{\pi}{3}}$ is a root of P(x), so that we can write...
$\displaystyle a_{5}\ e^{i\ \frac{5}{3}\ \pi} + a_{4}\ e^{i\ \frac{4}{3}\ \pi} + a_{3}\ e^{i\ \pi} + a_{2}\ e^{i\ \frac{2}{3}\ \pi} + a_{1}\ e^{i\ \frac{1}{3}\ \pi} + a_{0}= (a_{2}-a_{5})\ e^{i\ \frac{2}{3}\ \pi} + (a_{1}-a_{4})\ e^{i\ \frac{1}{3}\ \pi} + (a_{0}-a_{3})=0$ (1)
Observing (1) we note that all the triplets $\displaystyle a_{0},a_{1},a_{2}$ combined with the conditions $\displaystyle a_{3}=a_{0}, a_{4}=a_{1}, a_{5}=a_{2}$ satisfies (1) so that the requested number of polynomials is $\displaystyle 8^{3}= 512$...
If you take the real part of (1) then you find that $\frac12((a_{2}-a_{5}) +\frac12(a_{1}-a_{4}) + (a_{0}-a_{3}) = 0$. If you take the imaginary part of (1) then you get $a_{2}-a_{5} = a_{1}-a_{4}$. Those equations are equivalent to the boxed equations in my previous comment. These are the only conclusions that you can deduce from (1), and chisigma's following claim is incorrect:chisigma said:$\displaystyle a_{5}\ e^{i\ \frac{5}{3}\ \pi} + a_{4}\ e^{i\ \frac{4}{3}\ \pi} + a_{3}\ e^{i\ \pi} + a_{2}\ e^{i\ \frac{2}{3}\ \pi} + a_{1}\ e^{i\ \frac{1}{3}\ \pi} + a_{0}= (a_{2}-a_{5})\ e^{i\ \frac{2}{3}\ \pi} + (a_{1}-a_{4})\ e^{i\ \frac{1}{3}\ \pi} + (a_{0}-a_{3})=0$ (1)
chisigma said:Observing more carefully (1) we discover that the condition $\displaystyle a_{2}-a_{5}= a_{1}-a_{4}=a_{0}-a_{3}=0$ is not the only solving way because the basic equality $\displaystyle e^{i\ \frac{2}{3}\ \pi} - e^{i\ \frac{1}{3}\ \pi} + 1=0$ implies that $\displaystyle a_{2}-a_{5}= a_{4}-a_{1}=a_{0}-a_{3}= \pm 1$ is also a solving way and that means that the requested number of polynomials is $\displaystyle 8^{3} +2\ \cdot 7^{3}= 1198$...
I understood the phrase in red to mean that, in each such polynomial, the six integers $a_0,a_1,a_2,a_3,a_4,a_5$ should be distinct elements of the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$. That condition makes the problem very much harder to solve.jacks said:Find the number of polynomials of degree $5$ with distinct coefficients from the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$ that are divisible by $x^2 - x + 1$
Opalg said:If you take the real part of (1) then you find that $\frac12((a_{2}-a_{5}) +\frac12(a_{1}-a_{4}) + (a_{0}-a_{3}) = 0$. If you take the imaginary part of (1) then you get $a_{2}-a_{5} = a_{1}-a_{4}$. Those equations are equivalent to the boxed equations in my previous comment. These are the only conclusions that you can deduce from (1), and chisigma's following claim is incorrect:
The original problem was:
I understood the phrase in red to mean that, in each such polynomial, the six integers $a_0,a_1,a_2,a_3,a_4,a_5$ should be distinct elements of the set $\left\{1, 2, 3, 4, 5, 6, 7, 8\right\}$. That condition makes the problem very much harder to solve.
>function CountDiv()
$
$ global I;
$
$ count=0;
$ OKs=0;
$ r1=0.5*(1+sqrt(3)*I);
$ r2=0.5*(1-sqrt(3)*I);
$ arry1=1:8;
$
$ for idx1=1 to 7
$ for idx2=idx1+1 to 8
$ aa=nonzeros(arry1!=idx1);
$ arry2=arry1(aa);
$ aa=nonzeros(arry2!=idx2);
$ arry2=arry2(aa);
$
$ for jdx=1 to fak(6)-1
$ count=count+1;
$ c1=polyval(arry2,r1);
$ if c1~=0
$ c2=polyval(arry2,r2);
$ if c2~=0
$ OKs=OKs+1;
$ endif
$ endif
$ arry2=next(arry2);
$ end
$ end
$ end
$
$ return {count, OKs}
$ endfunction
>{C,O}=CountDiv;[C,O]
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