Found $x^8$ while calculating inverse of $(x^6+1)$ in finite field $GF(2^8)$. Help???Quadratic Extension of Finite fieldInverse of polynomial over $mathbb F_3$ finite field, quotient spaceExtended Euclidean Algorithm, what is our answer?Calculating the generator of a Finite FieldIs every finite extension of a finite field a finite field?Algebraic Field Extension of Finite FieldPrescribing norm and trace of elements in a finite field.Help determining if a field is finite?degree of finite extension of finite fieldHelp with a basic question on Finite Field characteristic

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Found $x^8$ while calculating inverse of $(x^6+1)$ in finite field $GF(2^8)$. Help???


Quadratic Extension of Finite fieldInverse of polynomial over $mathbb F_3$ finite field, quotient spaceExtended Euclidean Algorithm, what is our answer?Calculating the generator of a Finite FieldIs every finite extension of a finite field a finite field?Algebraic Field Extension of Finite FieldPrescribing norm and trace of elements in a finite field.Help determining if a field is finite?degree of finite extension of finite fieldHelp with a basic question on Finite Field characteristic













0












$begingroup$


So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got:
$$1=(x+1)-1(x)
=(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$

This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$
But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...



(EDIT)
enter image description here
This is an image of the EEA calculations










share|cite|improve this question









New contributor




hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$







  • 1




    $begingroup$
    Please use MathJax to make your question readable. Start by putting $ signs around the math expressions.
    $endgroup$
    – saulspatz
    2 days ago











  • $begingroup$
    Changed it, sorry didnt realize that at first..
    $endgroup$
    – hassan zaidi
    2 days ago










  • $begingroup$
    Welcome to Maths SX! What is $x$ here?
    $endgroup$
    – Bernard
    2 days ago










  • $begingroup$
    x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
    $endgroup$
    – hassan zaidi
    2 days ago










  • $begingroup$
    Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
    $endgroup$
    – hassan zaidi
    2 days ago















0












$begingroup$


So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got:
$$1=(x+1)-1(x)
=(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$

This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$
But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...



(EDIT)
enter image description here
This is an image of the EEA calculations










share|cite|improve this question









New contributor




hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 1




    $begingroup$
    Please use MathJax to make your question readable. Start by putting $ signs around the math expressions.
    $endgroup$
    – saulspatz
    2 days ago











  • $begingroup$
    Changed it, sorry didnt realize that at first..
    $endgroup$
    – hassan zaidi
    2 days ago










  • $begingroup$
    Welcome to Maths SX! What is $x$ here?
    $endgroup$
    – Bernard
    2 days ago










  • $begingroup$
    x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
    $endgroup$
    – hassan zaidi
    2 days ago










  • $begingroup$
    Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
    $endgroup$
    – hassan zaidi
    2 days ago













0












0








0


1



$begingroup$


So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got:
$$1=(x+1)-1(x)
=(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$

This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$
But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...



(EDIT)
enter image description here
This is an image of the EEA calculations










share|cite|improve this question









New contributor




hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got:
$$1=(x+1)-1(x)
=(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$

This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$
But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...



(EDIT)
enter image description here
This is an image of the EEA calculations







finite-fields extension-field euclidean-algorithm






share|cite|improve this question









New contributor




hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago







hassan zaidi













New contributor




hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









hassan zaidihassan zaidi

11




11




New contributor




hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 1




    $begingroup$
    Please use MathJax to make your question readable. Start by putting $ signs around the math expressions.
    $endgroup$
    – saulspatz
    2 days ago











  • $begingroup$
    Changed it, sorry didnt realize that at first..
    $endgroup$
    – hassan zaidi
    2 days ago










  • $begingroup$
    Welcome to Maths SX! What is $x$ here?
    $endgroup$
    – Bernard
    2 days ago










  • $begingroup$
    x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
    $endgroup$
    – hassan zaidi
    2 days ago










  • $begingroup$
    Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
    $endgroup$
    – hassan zaidi
    2 days ago












  • 1




    $begingroup$
    Please use MathJax to make your question readable. Start by putting $ signs around the math expressions.
    $endgroup$
    – saulspatz
    2 days ago











  • $begingroup$
    Changed it, sorry didnt realize that at first..
    $endgroup$
    – hassan zaidi
    2 days ago










  • $begingroup$
    Welcome to Maths SX! What is $x$ here?
    $endgroup$
    – Bernard
    2 days ago










  • $begingroup$
    x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
    $endgroup$
    – hassan zaidi
    2 days ago










  • $begingroup$
    Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
    $endgroup$
    – hassan zaidi
    2 days ago







1




1




$begingroup$
Please use MathJax to make your question readable. Start by putting $ signs around the math expressions.
$endgroup$
– saulspatz
2 days ago





$begingroup$
Please use MathJax to make your question readable. Start by putting $ signs around the math expressions.
$endgroup$
– saulspatz
2 days ago













$begingroup$
Changed it, sorry didnt realize that at first..
$endgroup$
– hassan zaidi
2 days ago




$begingroup$
Changed it, sorry didnt realize that at first..
$endgroup$
– hassan zaidi
2 days ago












$begingroup$
Welcome to Maths SX! What is $x$ here?
$endgroup$
– Bernard
2 days ago




$begingroup$
Welcome to Maths SX! What is $x$ here?
$endgroup$
– Bernard
2 days ago












$begingroup$
x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
$endgroup$
– hassan zaidi
2 days ago




$begingroup$
x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
$endgroup$
– hassan zaidi
2 days ago












$begingroup$
Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
$endgroup$
– hassan zaidi
2 days ago




$begingroup$
Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
$endgroup$
– hassan zaidi
2 days ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.



In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
    $endgroup$
    – hassan zaidi
    2 days ago











  • $begingroup$
    I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
    $endgroup$
    – lonza leggiera
    yesterday











  • $begingroup$
    Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
    $endgroup$
    – hassan zaidi
    yesterday










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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.



In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
    $endgroup$
    – hassan zaidi
    2 days ago











  • $begingroup$
    I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
    $endgroup$
    – lonza leggiera
    yesterday











  • $begingroup$
    Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
    $endgroup$
    – hassan zaidi
    yesterday















2












$begingroup$

Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.



In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
    $endgroup$
    – hassan zaidi
    2 days ago











  • $begingroup$
    I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
    $endgroup$
    – lonza leggiera
    yesterday











  • $begingroup$
    Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
    $endgroup$
    – hassan zaidi
    yesterday













2












2








2





$begingroup$

Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.



In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.






share|cite|improve this answer











$endgroup$



Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.



In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









lonza leggieralonza leggiera

1,03228




1,03228











  • $begingroup$
    Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
    $endgroup$
    – hassan zaidi
    2 days ago











  • $begingroup$
    I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
    $endgroup$
    – lonza leggiera
    yesterday











  • $begingroup$
    Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
    $endgroup$
    – hassan zaidi
    yesterday
















  • $begingroup$
    Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
    $endgroup$
    – hassan zaidi
    2 days ago











  • $begingroup$
    I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
    $endgroup$
    – lonza leggiera
    yesterday











  • $begingroup$
    Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
    $endgroup$
    – hassan zaidi
    yesterday















$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
$endgroup$
– hassan zaidi
2 days ago





$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
$endgroup$
– hassan zaidi
2 days ago













$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
$endgroup$
– lonza leggiera
yesterday





$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
$endgroup$
– lonza leggiera
yesterday













$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
$endgroup$
– hassan zaidi
yesterday




$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
$endgroup$
– hassan zaidi
yesterday










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