How to Factorize :
$$f(n)=a^n+a±1:n\in \mathbb{N}$$
my try:
$f(n)=a^n+a±1:n\in \mathbb{N} \to f(n)=a^n+a^{n-2}-a^{n-2}+a+1\\f(n)=a^{n-2}(a^2+a^{3-n})-a^{n-2}+1!!!!:($
How to factorize $f(n)=a^n+a±1$
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$\begingroup$
algebra-precalculus
polynomials
factoring
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0Who said it's possible in the first place? – 2017-01-18
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0It may not be as simple as we think. – 2017-01-18
2 Answers
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There is factorization of $a^n+a+1$ for all $n=3k+2$, where $k\in\mathbb N$ and of $a^n+a-1$ for all $n=3k+2$, where $k\in\mathbb 2N-1$.
Because we have $a^3+1=(a^2-a+1)(a+1)$ and $a^3-1=(a^2+a+1)(a-1)$. For example, $$a^5+a-1=a^5+a^2-a^2+a-1=(a^2-a+1)(a^3+a^2-1)$$
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0In general, what is? – 2017-01-18
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0@Almot1960 In the general case you can't do it in $\mathbb Z[a]$. Take $a^2+a+1$ or $a^2+a-1$. – 2017-01-18
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In general you cannot. For example the polynomial $x^5-x+1$ is not factorizable. See Abel- Ruffini theorem which says there is no resolution formula by radicals to solve a n-th degree polynomial for$\ 5\leq n $.