Simplify into reduced form

Question

Simplify into reduced form the expression $$\frac{1+2α+3α^2} {(1+α)^{15}} $$ , where the minimal polynomial of $α ∈ \mathbb{C}$ over $\mathbb{Q}$ is $x^3 + x + 1$.

My Attempt: First, we can write $\frac{-1}{\alpha} = 1 + \alpha^2$, thus, we can evaluate $(1+2\alpha + 3\alpha^2)(1+\alpha^2)^{45}$. I tried to compute and find a pattern for $(1+\alpha^2)^n$, but nothing! The final answer based on Computer Algebra Systems is not at all simple. Could there be a typo in the given expression?

Thank you for your time.

• Ghoster: I tried to compute and find a pattern... Use repeated squaring to calculate $(1+\alpha^2)^n$ for $n=1,2,4,8,16,32$. Then use the results for $n=1,4,8,32$ to calculate it for $n=45$.

Answer 1

The final answer based on Computer Algebra Systems is not at all simple.

Here is a finite upper bound on the length of a possible solution:

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The equation $\alpha^3+\alpha+1=0$ can be rewritten $\alpha^3=-\left(1+\alpha\right)$, $\frac{1}{\alpha}=-\left(1+\alpha^2\right)$; then

$$ \begin{align} \left(1+\alpha^2\right)^5 &= 1+5\cdot\alpha^2+10\cdot\alpha^4+10\cdot\alpha^6+5\cdot\alpha^8+\alpha^{10} =\\ &= 1+5\cdot\alpha^2+\alpha^3\cdot\left(10\cdot\alpha+\alpha^3\cdot\left(10+5\cdot\alpha^2+\alpha^3\cdot\alpha\right)\right) =\\ &= 1+5\cdot\alpha^2-\left(1+\alpha\right)\cdot\left(10\cdot\alpha-\left(1+\alpha\right)\cdot\left(10+5\cdot\alpha^2-\left(1+\alpha\right)\cdot\alpha\right)\right) =\\ &= 11+9\cdot\alpha+7\cdot\alpha^2+\alpha^3\cdot\left(7+4\cdot\alpha\right) =\\ &= 11+9\cdot\alpha+7\cdot\alpha^2-\left(1+\alpha\right)\cdot\left(7+4\cdot\alpha\right) =\\ &= 4-2\cdot\alpha+3\cdot\alpha^2 \\ \left(1+\alpha^2\right)^6 &= \left(1+\alpha^2\right)\cdot\left(4-2\cdot\alpha+3\cdot\alpha^2\right) =\\ &= 4-2\cdot\alpha+7\cdot\alpha^2-\alpha^3\cdot\left(2-3\cdot\alpha\right) =\\ &= 4-2\cdot\alpha+7\cdot\alpha^2+\left(1+\alpha\right)\cdot\left(2-3\cdot\alpha\right) =\\ &= 6-3\cdot\alpha+4\cdot\alpha^2 \end{align} $$

$$ \begin{align} \left(1+\alpha^2\right)^{36} &= \left(6-3\cdot\alpha+4\cdot\alpha^2\right)^6 =\\ &= \left(36-36\cdot\alpha+57\cdot\alpha^2-\alpha^3\cdot\left(24-16\cdot\alpha\right)\right)^3 =\\ &= \left(36-36\cdot\alpha+57\cdot\alpha^2+\left(1+\alpha\right)\cdot\left(24-16\cdot\alpha\right)\right)^3 =\\ &= \left(60-28\cdot\alpha+41\cdot\alpha^2\right)^3 =\\ &= \left(60-28\cdot\alpha+41\cdot\alpha^2\right)\cdot\left(3600-3360\cdot\alpha+5704\cdot\alpha^2-\alpha^3\cdot\left(2296-1681\cdot\alpha\right)\right) =\\ &= \left(60-28\cdot\alpha+41\cdot\alpha^2\right)\cdot\left(3600-3360\cdot\alpha+5704\cdot\alpha^2+\left(1+\alpha\right)\cdot\left(2296-1681\cdot\alpha\right)\right) =\\ &= \left(60-28\cdot\alpha+41\cdot\alpha^2\right)\cdot\left(5896-2745\cdot\alpha+4023\cdot\alpha^2\right) =\\ &= 353760-329788\cdot\alpha+559976\cdot\alpha^2-\alpha^3\cdot\left(225189-164943\cdot\alpha\right) =\\ &= 353760-329788\cdot\alpha+559976\cdot\alpha^2+\left(1+\alpha\right)\cdot\left(225189-164943\cdot\alpha\right) =\\ &= 578949-269542\cdot\alpha+395033\cdot\alpha^2 \end{align} $$

$$ \begin{align} x &= \frac{1+2\cdot\alpha+3\cdot\alpha^2}{\left(1+\alpha\right)^{15}} =\\ &= \frac{\left(1+\alpha^2\right)+2\cdot\alpha\cdot\left(1+\alpha\right)}{\left(-\alpha^3\right)^{15}} =\\ &= \frac{-\frac{1}{\alpha}-2\cdot\alpha\cdot\alpha^3}{-\alpha^{45}} = \frac{1}{\alpha^{46}}+\frac{2}{\alpha^{41}} =\\ &= \left(1+\alpha^2\right)^{46}-2\cdot\left(1+\alpha^2\right)^{41} =\\ &= \left(\left(1+\alpha^2\right)^5-2\right)\cdot\left(1+\alpha^2\right)^5\cdot\left(1+\alpha^2\right)^{36} =\\ &= \left(2-2\cdot\alpha+3\cdot\alpha^2\right)\cdot\left(4-2\cdot\alpha+3\cdot\alpha^2\right)\cdot\left(578949-269542\cdot\alpha+395033\cdot\alpha^2\right) =\\ &= \left(8-12\cdot\alpha+22\cdot\alpha^2-\alpha^3\cdot\left(12-9\cdot\alpha\right)\right)\cdot\left(578949-269542\cdot\alpha+395033\cdot\alpha^2\right) =\\ &= \left(20-9\cdot\alpha+13\cdot\alpha^2\right)\cdot\left(578949-269542\cdot\alpha+395033\cdot\alpha^2\right) =\\ &= 11578980-10601381\cdot\alpha+17852875\cdot\alpha^2-\alpha^3\cdot\left(7059343-5135429\cdot\alpha\right) =\\ &= 11578980-10601381\cdot\alpha+17852875\cdot\alpha^2+\left(1+\alpha\right)\cdot\left(7059343-5135429\cdot\alpha\right) =\\ &= 18638323-8677467\cdot\alpha+12717446\cdot\alpha^2 \end{align} $$

Answer 2

With a professional tool (Pari/GP), this is purely mechanical:

(15:46) gp > a=Mod(x,x^3+x+1)
%1 = Mod(x, x^3 + x + 1)
(15:47) gp > (1+2*a+3*a^2)/(1+a)^15
%2 = Mod(12717446*x^2 - 8677467*x + 18638323, x^3 + x + 1)

Of course, this is in agreement with the result of @mezzoctane.

Answer 3

1ドル+2a+3a^2 = (1+a)^2+2a^2$, and $a+1 = -a^3$.

So $\frac{1+2a+3a^2}{(1+a)^{15}} = \frac{-a^3+2a^2}{(-a^3)^{15}} = a^{-42}-2a^{-43}$.

• user21820: I'm not sure what "reduced form" is intended to mean, but I guess that this is the simplest possible expression for the answer...