What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$? [duplicate]
This question already has an answer here:
Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$
2 answers
What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$?
Is there any open softwares to calculate such things easily?
ring-theory inverse irreducible-polynomials
marked as duplicate by Saad, GNUSupporter 8964民主女神 地下教會, jgon, Bill Dubuque
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Dec 4 '18 at 15:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$
2 answers
What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$?
Is there any open softwares to calculate such things easily?
ring-theory inverse irreducible-polynomials
marked as duplicate by Saad, GNUSupporter 8964民主女神 地下教會, jgon, Bill Dubuque
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Dec 4 '18 at 15:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Hint $bmod, 1+xf!:, (-f)xequiv 1 $
– Bill Dubuque
Dec 4 '18 at 14:40
add a comment |
This question already has an answer here:
Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$
2 answers
What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$?
Is there any open softwares to calculate such things easily?
ring-theory inverse irreducible-polynomials
This question already has an answer here:
Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$
2 answers
What is the inverse of X modulo $1 + X + X^2 + X^3 + X^4$?
Is there any open softwares to calculate such things easily?
This question already has an answer here:
Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$
2 answers
ring-theory inverse irreducible-polynomials
ring-theory inverse irreducible-polynomials
edited Dec 4 '18 at 9:26
José Carlos Santos
152k22123225
152k22123225
asked Dec 4 '18 at 9:23
malleamallea
29919
29919
marked as duplicate by Saad, GNUSupporter 8964民主女神 地下教會, jgon, Bill Dubuque
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Dec 4 '18 at 15:25
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Dec 4 '18 at 15:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Hint $bmod, 1+xf!:, (-f)xequiv 1 $
– Bill Dubuque
Dec 4 '18 at 14:40
add a comment |
1
Hint $bmod, 1+xf!:, (-f)xequiv 1 $
– Bill Dubuque
Dec 4 '18 at 14:40
1
1
Hint $bmod, 1+xf!:, (-f)xequiv 1 $
– Bill Dubuque
Dec 4 '18 at 14:40
Hint $bmod, 1+xf!:, (-f)xequiv 1 $
– Bill Dubuque
Dec 4 '18 at 14:40
add a comment |
1 Answer
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Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.
Got it! Thank you very much :)
– mallea
Dec 4 '18 at 13:33
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1 Answer
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1 Answer
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active
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Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.
Got it! Thank you very much :)
– mallea
Dec 4 '18 at 13:33
add a comment |
Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.
Got it! Thank you very much :)
– mallea
Dec 4 '18 at 13:33
add a comment |
Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.
Hint: $1+X+X^2+X^3+X^4=1+X(1+X+X^2+X^3)$.
answered Dec 4 '18 at 9:25
José Carlos SantosJosé Carlos Santos
152k22123225
152k22123225
Got it! Thank you very much :)
– mallea
Dec 4 '18 at 13:33
add a comment |
Got it! Thank you very much :)
– mallea
Dec 4 '18 at 13:33
Got it! Thank you very much :)
– mallea
Dec 4 '18 at 13:33
Got it! Thank you very much :)
– mallea
Dec 4 '18 at 13:33
add a comment |
1
Hint $bmod, 1+xf!:, (-f)xequiv 1 $
– Bill Dubuque
Dec 4 '18 at 14:40