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Will Jagy
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NOTE: I am doing $x^4 + 12 x^3 + 14 x^2 - 12 x + 1.$

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10},$ after $$(x+1)^4 \pmod 2, \; \; (x^2 + 1)^2 \pmod 3, \; \; (x+3)^4 \pmod 5. $$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10},$ after $$(x+1)^4 \pmod 2, \; \; (x^2 + 1)^2 \pmod 3, \; \; (x+3)^4 \pmod 5. $$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$

NOTE: I am doing $x^4 + 12 x^3 + 14 x^2 - 12 x + 1.$

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10},$ after $$(x+1)^4 \pmod 2, \; \; (x^2 + 1)^2 \pmod 3, \; \; (x+3)^4 \pmod 5. $$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$

added 78 characters in body
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Will Jagy
  • 25.7k
  • 2
  • 65
  • 121

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10}.$$p \equiv \pm 3 \pmod {10},$ after $$(x+1)^4 \pmod 2, \; \; (x^2 + 1)^2 \pmod 3, \; \; (x+3)^4 \pmod 5. $$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10}.$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10},$ after $$(x+1)^4 \pmod 2, \; \; (x^2 + 1)^2 \pmod 3, \; \; (x+3)^4 \pmod 5. $$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$

Source Link
Will Jagy
  • 25.7k
  • 2
  • 65
  • 121

Something is very wrong, perhaps just the typing of the polynomial. With gp-pari, I do get irreducible $\pmod p$ for primes $p \equiv \pm 3 \pmod {10}.$

Pari says the discriminant is $2^{12} \cdot 3^2 \cdot 5^3.$

It gives four linear factors for $$ p \in \{ 29, 31, 59, 61, 89, 149, 151, 179, 181, 211, 239, 241, \ldots \} $$

It gives two quadratic factors for $$ p \in \{ 11,19,41,71,79,101,109,131,139,191,199,229,251, \ldots, 409,\ldots \} $$