{"id":358,"date":"2026-09-21T18:32:28","date_gmt":"2026-09-21T18:32:28","guid":{"rendered":"https:\/\/mathrelated.co.uk\/?p=358"},"modified":"2026-09-22T18:43:48","modified_gmt":"2026-09-22T18:43:48","slug":"proving-proth-number-prime-using-euler-extra-step","status":"publish","type":"post","link":"https:\/\/mathrelated.co.uk\/index.php\/2026\/09\/21\/proving-proth-number-prime-using-euler-extra-step\/","title":{"rendered":"Proving Proth number prime using Euler – extra step"},"content":{"rendered":"\n

In the previous post we offered a partial answer to proving a Proth Number is prime<\/p>\n\n\n\n

The result would be a probable prime to base 2.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

.<\/p>\n\n\n\n

\"Table<\/figure>\n\n\n\n

74665 is not a Proth Number. Is 74665 a strong Euler-Jacobi Pseudoprime?<\/p>\n\n\n\n

Of the 37 initial terms in OEIS sequence A047713<\/a> we found the following:<\/p>\n\n\n\n

One Pseudoprime passed the filter – 74665<\/p>\n\n\n\n

Three were Incalculable as p-1 not divisible by 4<\/p>\n\n\n\n

Five were marked ‘not applicable’ as we can only apply our filter when (2\/p)=1<\/p>\n\n\n\n

One in twenty-nine passed the filter which means we eliminated 96% of the Pseudoprimes using this extra filter.<\/p>\n\n\n\n

If you count the 8 for which our filter could not be used then we still eliminated 3\/4 of the Pseudoprimes.<\/p>\n\n\n\n

For the conjecture at the top of this page, we would always be dealing with (p-1) divisible by 8 and 2^((p-1)\/2) = 1 mod p<\/p>\n\n\n\n

So our filter would likely achieve near the 96% we achieved here.<\/p>\n\n\n\n

Having amended the previous post, this filter is already incorporated into our prime test.<\/p>\n\n\n\n

If reading this post has got you interested in Probable Prime theory then the following resources might offer some further reading:<\/p>\n\n\n\n