## Algebraic identity for the 4th degree Aurifeuillian algebraic factorization

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• I followed the links given by David, and taking the hints given, constructed this identity: m**2 A**4 - m * (m-2) A**2 B**2 + B**4 = ( m * A**2 - m * A * B +
Message 1 of 2 , Feb 3, 2011
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I followed the links given by David, and taking the hints given,

constructed this identity:

m**2 A**4 - m * (m-2) A**2 B**2 + B**4

= ( m * A**2 - m * A * B + B**2 )
* ( m * A**2 + m * A * B + B**2)

Now the only puzzle remaining is how to match up

the form

m**2 A**4 - m * (m-2) A**2 B**2 + B**4

to the number you wish to factor.

Kermit
• ... That 4th degree equation is useful (in the Aurifeullian sense) only at m=2. As I explained, the general Aurifeullian identity for Phi(n,x) requires (among
Message 2 of 2 , Feb 3, 2011
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Kermit Rose <kermit@...> wrote:

> m**2 A**4 - m * (m-2) A**2 B**2 + B**4
> = ( m * A**2 - m * A * B + B**2 )
> * ( m * A**2 + m * A * B + B**2)

That 4th degree equation is useful (in the
Aurifeullian sense) only at m=2. As I explained,
the general Aurifeullian identity for Phi(n,x)
requires (among other things) that x = m*s^2,
where m divides n. For example:

Phi(n,b)=subst(polcyclo(n),x,b);

print(factor(Phi(4,2*x^2))[,1]~)
[2*x^2 - 2*x + 1, 2*x^2 + 2*x + 1]

print(factor(Phi(6,3*x^2))[,1]~)
[3*x^2 - 3*x + 1, 3*x^2 + 3*x + 1]

print(factor(Phi(5,5*x^2))[,1]~)
[25*x^4 - 25*x^3 + 15*x^2 - 5*x + 1,
25*x^4 + 25*x^3 + 15*x^2 + 5*x + 1]

print(factor(Phi(15,5*x^2))[,1]~)
[625*x^8 - 625*x^7 + 250*x^6 - 125*x^5 + 75*x^4 - 25*x^3 + 10*x^2 - 5*x + 1,
625*x^8 + 625*x^7 + 250*x^6 + 125*x^5 + 75*x^4 + 25*x^3 + 10*x^2 + 5*x + 1]

print(factor(Phi(30,3*x^2))[,1]~)
[81*x^8 - 81*x^7 + 54*x^6 - 27*x^5 + 9*x^4 - 9*x^3 + 6*x^2 - 3*x + 1,
81*x^8 + 81*x^7 + 54*x^6 + 27*x^5 + 9*x^4 + 9*x^3 + 6*x^2 + 3*x + 1]

David
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