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January 8

Even Fermat pseudoprimes

There are infinitely many odd numbers which are Fermat pseudoprimes to all bases coprime to them (the Carmichael numbers), but it seems that all even numbers > 946 are Fermat pseudoprimes to at most 1/8 for the bases coprime to them, is this proven? (Like that all odd numbers are strong pseudoprimes to at most 1/4 for the bases coprime to them) 220.132.216.52 (talk) 12:40, 8 January 2025 (UTC)

For what it's worth, any counterexample >28 must have the form $2pq$ where p and q are distinct odd primes such that $2pq-1$ is divisible by both $(p-1)/2$ and $(q-1)/2$. Tito Omburo (talk) 18:55, 8 January 2025 (UTC)

If we denote $a:=\gcd ((p-1)/2,(q-1)/2)$ such that $P:=(p-1)/2a$ and $Q:=(q-1)/2a$ are coprime, then $(p-1)/2=aP$ and $(q-1)/2=aQ$ divides $2pq-1=2(2aP+1)(2aQ+1)-1=8a^{2}PQ+4aP+4aQ+1$. Naturally, this implies $aP$ divides $4aQ+1$ and $aQ$ divides $4aP+1$, which further implies that $a=1$ and thus $(p-1)/2$ and $(q-1)/2$ are coprime. So we must find (odd) coprime $P,Q$ such that $P|4Q+1$ and $Q|4P+1$, and from there $2P+1$ and $2Q+1$ must be prime. GalacticShoe (talk) 16:13, 16 January 2025 (UTC)

Suppose we have odd coprime $P,Q$ such that $P|4Q+1$ and $Q|4P+1$. Assume WLOG that $P<Q$. This means that $4P+1$ can only be $Q$ or $3Q$. If $4P+1=Q$, then $P|4Q+1\Rightarrow P|4(4P+1)+1\Rightarrow P|16P+5\Rightarrow P=1,5,Q=5,21$. This yields $p:=2P+1=3,11,q:=2Q+1=11,43$ which both work, giving values $2pq=66,946$. If $4P+1=3Q$, then $P|4(4P+1)/3+1\Rightarrow P|(16P+7)/3\Rightarrow P|16P+7\Rightarrow P=1,7$. For either value though $4P+1$ is not divisible by $3$, so this doesn't yield any values. We conclude that the only counterexamples greater than $28$ are indeed $66,946$. GalacticShoe (talk) 01:32, 17 January 2025 (UTC)

It is notable that 28, 66, 946 are triangular numbers, and their indices (7, 11, 43) are Heegner numbers, is this a coincidence? 220.132.216.52 (talk) 10:59, 17 January 2025 (UTC)

The other triangular numbers whose indices are Heegner numbers are 190 (19), 2278 (67), 13366 (163), but 190 is Fermat pseudoprime only to 1/8 for the bases coprime to it, 2278 is Fermat pseudoprime only to 1/32 for the bases coprime to it, 13366 is Fermat pseudoprime only to 1/16 for the bases coprime to it (8, 32, 16 are powers of 2). 220.132.216.52 (talk) 13:09, 17 January 2025 (UTC)

I seem to know the reason:

(Heegner numbers corresponding to the prime-generating polynomial n^2+n+p, i.e. the number p = (the Heegner number + 1)/4

66 -> 11th triangular number -> 11 and (11+1)/4 = 3 -> 11-1 and 3-1 totally have 2 prime factors of 2 -> 1/(2^2) = 1/4 of the bases coprime to it

190 -> 19th triangular number -> 19 and (19+1)/4 = 5 -> 19-1 and 5-1 totally have 3 prime factors of 2 -> 1/(2^3) = 1/8 of the bases coprime to it

946 -> 43rd triangular number -> 43 and (43+1)/4 = 11 -> 43-1 and 11-1 totally have 2 prime factors of 2 -> 1/(2^2) = 1/4 of the bases coprime to it

2278 -> 67th triangular number -> 67 and (67+1)/4 = 17 -> 67-1 and 17-1 totally have 5 prime factors of 2 -> 1/(2^5) = 1/32 of the bases coprime to it

13366 -> 163rd triangular number -> 163 and (163+1)/4 = 41 -> 163-1 and 41-1 totally have 4 prime factors of 2 -> 1/(2^4) = 1/16 of the bases coprime to it 220.132.216.52 (talk) 13:16, 17 January 2025 (UTC)

Template:Ping Template:Ping Template:Ping Template:Ping 220.132.216.52 (talk) 14:45, 16 January 2025 (UTC)

Rectifying latitude

Is it true that using the rectifying latitude with the best sphere radius (possibly not the same radius for all 3) minimizes the worst-case error for distance (%), distance (km) & max km the 2 paths get from each other? (maximum separation between the great circle & the geodesic for the surface of the WGS84 ellipsoid)? Or is another latitude better like the geocentric latitude? (the geocentric latitude can get ~0.2° from the (by far) most kind of used latitude which is more separation than any kind of latitude (besides the Mercator one that's 0° to ∞°)) What's the best radius for each of these 3 metrics & how well do the worst point pairs for these 3 metrics approximate the ellipsoidal trigonometry answer? (the one where the geodesic latA lon A alt0 to latB lonB alt0 is considered perfect accuracy even though most places aren't on the 2D surface) Sagittarian Milky Way (talk) 22:56, 8 January 2025 (UTC)