5

Let $P(x) = x^3 + 2x^2+3x+4$ and $a$ be the root of equation $x^4+x^3+x^2+x+1=0$.

Find the value of $P(a)P(a^2)P(a^3)P(a^4)$

Is my answer correct ?

Since root of equation $x^4+x^3+x^2+x+1=0$ is the $5^{th}$ primitive root of 1,

so $a, a^2, a^3, a^4$ are roots of $x^4+x^3+x^2+x+1=0$

$P(a)P(a^4)=(a^3+2a^2+3a+4)(\frac{1}{a^3}+\frac{2}{a^2}+\frac{3}{a}+4)= 15+5a^4+5a$

Similarly, $P(a^2)P(a^3)=15+5a^3+5a^2$

$P(a)P(a^2)P(a^3)P(a^4)=(15+5a^4+5a)(15+5a^3+5a^2)=125$

user403160
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  • First of all, I'm guessing that you meant $a = \operatorname{cis}(2\pi/5)$, and secondly, you cannot assume that $a = \operatorname{cis}(2\pi/5)$ a priori. However, it is still true that one of $a,a^2,a^3,a^4$ is $ \operatorname{cis}(2\pi/5)$ and more importantly, they are all distinct. In the end, sure, you can say that $a = \operatorname{cis}(2\pi/5)$ is not loss of generality. – Ennar Aug 22 '17 at 14:49
  • I think that since $a$ and $a^4$ are complex conjugates $P(a)P(a^4)$ should be real, so there is a glitch in your calculation. I have sketched out a solution, which I'm not posting yet, based on looking at $(x-1)P(x)$. This gives me an integer answer related (as one might expect) to the prime $5$. – Mark Bennet Aug 22 '17 at 14:59
  • I've edited my answer. please check if it's correct or not. – user403160 Aug 22 '17 at 15:01

2 Answers2

5

Suppose $P(x)=x^3+2x^2+3x+4=(x-p)(x-q)(x-r)$ for some $p,q,r\in\mathbb{C}$. Then, $$\prod_{j=1}^4\,P\left(a^j\right)=Q(p)\,Q(q)\,Q(r)\,,$$ where $Q(x):=x^4+x^3+x^2+x+1$. Now, $$Q(x)=(x-1)\,P(x)+5\,.$$ Thus, $Q(p)=Q(q)=Q(r)=5$.

This is actually quite a nice technique. Let $P(x)$ and $Q(x)$ be two nonconstant polynomials in $x$ over a field $K$. Write $\bar{P}$ and $\bar{Q}$ for the leading coefficients of $P(x)$ and $Q(x)$, respectively. Suppose that $t_1,t_2,\ldots,t_n$ are the roots of $Q(x)$ in the algebraic closure of $K$ (with multiplicities). Then, $$\prod_{j=1}^n\,P\left(t_j\right)=(-1)^{mn}\,\frac{\bar{P}^n}{\bar{Q}^m}\,\prod_{i=1}^m\,Q\left(s_i\right)\,,\tag{1}$$ where $s_1,s_2,\ldots,s_m$ are the roots of $P(x)$in the algebraic closure of $K$ (with multiplicities).

If $m>n$, then write $$P(x)=Q(x)\,A(x)+B(x)$$ for some $A(x),B(x)\in K[x]$ with $B(x)$ having degree less than $n$. Clearly, we have $$\prod_{j=1}^n\,P\left(t_j\right)=\prod_{i=1}^m\,B\left(t_i\right)\,,\tag{2}$$ Then, we can use the paragraph below using $B(x)$ in place of $P(x)$ to simplify things even more, provided that $B(x)$ is nonconstant (or nonlinear).

If $m \leq n$, then we write $Q(x)=P(x)\,U(x)+V(x)$ for some polynomials $U(x),V(x)\in K[x]$, with $V(x)$ having degree less than $m$. Then, from $(1)$, $$\prod_{j=1}^n\,P\left(t_j\right)=(-1)^{mn}\,\frac{\bar{P}^n}{\bar{Q}^m}\,\prod_{i=1}^m\,V\left(s_i\right)\,.\tag{3}$$ In fact, we can play this game again between $V(x)$ and $P(x)$ to make further degree reductions, provided that $V(x)$ is nonconstant (or nonlinear).

For example, let $P(x)=x^5+x^2+2x+1$ and $Q(x)=x^3-3x-2$. Then, we see that $P(x)=A(x)\,Q(x)+B(x)$ with $A(x)=x^2+3$ and $B(x)=3x^2+11x+7$. If $t_1$, $t_2$, and $t_3$ are the roots of $Q(x)$, then $$\prod_{j=1}^3\,P\left(t_j\right)=\prod_{j=1}^3\,B\left(t_j\right)\,.$$ Note that $Q(x)=B(x)\,U(x)+V(x)$, where $U(x)=\frac{x}{3}-\frac{11}{9}$ and $V(x)=\frac{73}{9}x+\frac{59}{9}$. Using $(3)$, we obtain $$\prod_{j=1}^3\,B\left(t_j\right)=(-1)^{2\cdot 3}\,\frac{3^3}{1^2}\,\prod_{i=1}^2\,V\left(r_i\right)\,,$$ where $r_1$ and $r_2$ are the roots of $B(x)$. That is, $$\prod_{j=1}^3\,P\left(t_j\right)=3^3\,\prod_{i=1}^2\,V\left(r_i\right)=3^2\left(\frac{73}{9}\right)^2\left(-\frac{59}{73}-r_1\right)\left(-\frac{59}{73}-r_2\right)=3^2\left(\frac{73}{9}\right)^2\,B\left(-\frac{59}{73}\right)\,.$$ That is, $\prod_{j=1}^3\,P\left(t_j\right)=41$. This can be easily verified as $t_1=-1$, $t_2=-1$, and $t_3=2$, so that $P\left(t_1\right)=-1$, $P\left(t_2\right)=-1$, and $P\left(t_3\right)=41$. (I picked an easy-to-check example, of course, so this reduction method is actually more complicated than simply computing the values of $P\left(t_j\right)$'s.)

Batominovski
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1

Here is another way. Call the product that you want Q.

Note that $xP(x)-P(x)=x^4+x^3+x^2+x-4$

As you correctly observe the values you are substituting are all roots of $x^4+x^3+x^2+x+1=0$ and hence for these values you get $(x-1)P(x)=-5$ whence $$(a-1)(a^2-1)(a^3-1)(a^4-1)Q=(-5)^4=625$$

Now pairing the first and last factor and the middle two factors and noting that $a^5=1$ we have $$(a-1)(a^2-1)(a^3-1)(a^4-1)=(2-a-a^4)(2-a^2-a^3) =$$$$=4-2(a+a^2+a^3+a^4)+a^3+a^4+a+a^2$$ and we know that $$a^4+a^3+a^2+a=-1$$ So $5Q=625$ and $Q=125$

Mark Bennet
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    Alternatively, noting that $a,a^2,a^3,a^4$ are roots of $x^4+x^3+x^2+x+1=0$ we can set $y=x-1$ and obtain $(y+1)^4+(y+1)^3+(y+1)^2+(y+1)+1=0$ as the polynomial satisfied by $a-1, a^2-1, a^3-1, a^4-1$. We want the product of the roots, which is the constant term, which is $5$. At least the calculation checks. – Mark Bennet Aug 22 '17 at 15:37
  • Your solution is easier than mine. Thank you for your help, Mark Bennet. – user403160 Aug 22 '17 at 15:38