lucas primality test small primes

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I'm writing a a Lucas primality test using the Selfridge parameters, but it seems to indicate that 5 and 11 are not prime. For both of those, the Jacobi number is 0.



In the case of 5, the first Jacobi test you try is (5/5) = 0, so you stop and return composite.



In the case of 11, you do (5/11) = 1, (-7/11) = 1, (9/11) = 1, (-11/11) = 0.



I found no other examples so far where my test returns it's composite while it's prime.



They both stop when |D| = n. I can't find any indication that I should stop searching for D when D reaches some value, only that I should check that it's not a perfect square.



Is |D| = n some special case? Should I continue searching for D in that case? At least 5 and 11 will find -1 for the Jacobi symbol for the next value in the sequence.




primality-test




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  • I added a "if (j != 1 && |D| != n) break" in my loop, meaning we want to stop searching if the Jacobi result is 0 or -1, but skip the case where the Jacobi test is not meaningful. Optionally I could skip the Jacobi test entirely but that's a dubious micro-optimization. Realistically this only happens for 5 and 11.
    – DanaJ
    2 days ago














up vote
-1
down vote

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I'm writing a a Lucas primality test using the Selfridge parameters, but it seems to indicate that 5 and 11 are not prime. For both of those, the Jacobi number is 0.



In the case of 5, the first Jacobi test you try is (5/5) = 0, so you stop and return composite.



In the case of 11, you do (5/11) = 1, (-7/11) = 1, (9/11) = 1, (-11/11) = 0.



I found no other examples so far where my test returns it's composite while it's prime.



They both stop when |D| = n. I can't find any indication that I should stop searching for D when D reaches some value, only that I should check that it's not a perfect square.



Is |D| = n some special case? Should I continue searching for D in that case? At least 5 and 11 will find -1 for the Jacobi symbol for the next value in the sequence.




primality-test




share|cite|improve this question



















  • I added a "if (j != 1 && |D| != n) break" in my loop, meaning we want to stop searching if the Jacobi result is 0 or -1, but skip the case where the Jacobi test is not meaningful. Optionally I could skip the Jacobi test entirely but that's a dubious micro-optimization. Realistically this only happens for 5 and 11.
    – DanaJ
    2 days ago












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I'm writing a a Lucas primality test using the Selfridge parameters, but it seems to indicate that 5 and 11 are not prime. For both of those, the Jacobi number is 0.



In the case of 5, the first Jacobi test you try is (5/5) = 0, so you stop and return composite.



In the case of 11, you do (5/11) = 1, (-7/11) = 1, (9/11) = 1, (-11/11) = 0.



I found no other examples so far where my test returns it's composite while it's prime.



They both stop when |D| = n. I can't find any indication that I should stop searching for D when D reaches some value, only that I should check that it's not a perfect square.



Is |D| = n some special case? Should I continue searching for D in that case? At least 5 and 11 will find -1 for the Jacobi symbol for the next value in the sequence.




primality-test




share|cite|improve this question











I'm writing a a Lucas primality test using the Selfridge parameters, but it seems to indicate that 5 and 11 are not prime. For both of those, the Jacobi number is 0.



In the case of 5, the first Jacobi test you try is (5/5) = 0, so you stop and return composite.



In the case of 11, you do (5/11) = 1, (-7/11) = 1, (9/11) = 1, (-11/11) = 0.



I found no other examples so far where my test returns it's composite while it's prime.



They both stop when |D| = n. I can't find any indication that I should stop searching for D when D reaches some value, only that I should check that it's not a perfect square.



Is |D| = n some special case? Should I continue searching for D in that case? At least 5 and 11 will find -1 for the Jacobi symbol for the next value in the sequence.





primality-test





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asked Jul 30 at 4:44









Kurt Roeckx

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  • I added a "if (j != 1 && |D| != n) break" in my loop, meaning we want to stop searching if the Jacobi result is 0 or -1, but skip the case where the Jacobi test is not meaningful. Optionally I could skip the Jacobi test entirely but that's a dubious micro-optimization. Realistically this only happens for 5 and 11.
    – DanaJ
    2 days ago
















  • I added a "if (j != 1 && |D| != n) break" in my loop, meaning we want to stop searching if the Jacobi result is 0 or -1, but skip the case where the Jacobi test is not meaningful. Optionally I could skip the Jacobi test entirely but that's a dubious micro-optimization. Realistically this only happens for 5 and 11.
    – DanaJ
    2 days ago















I added a "if (j != 1 && |D| != n) break" in my loop, meaning we want to stop searching if the Jacobi result is 0 or -1, but skip the case where the Jacobi test is not meaningful. Optionally I could skip the Jacobi test entirely but that's a dubious micro-optimization. Realistically this only happens for 5 and 11.
– DanaJ
2 days ago




I added a "if (j != 1 && |D| != n) break" in my loop, meaning we want to stop searching if the Jacobi result is 0 or -1, but skip the case where the Jacobi test is not meaningful. Optionally I could skip the Jacobi test entirely but that's a dubious micro-optimization. Realistically this only happens for 5 and 11.
– DanaJ
2 days ago















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