The discussion revolves around finding 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 scope includes mathematical reasoning and exploration of conditions related to polynomial coefficients.
Participants express differing views on the interpretation of "distinct coefficients" and its implications for the problem. There is no consensus on the final count of polynomials, with multiple competing models and calculations presented throughout the discussion.
Participants note that the problem's complexity arises from the requirement for distinct coefficients and the conditions derived from the polynomial's divisibility. There are unresolved assumptions regarding the interpretation of the problem statement and the implications for counting valid polynomials.
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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