[QUOTE=science_man_88;483253]. Not really, it happens any time both bases are squares.[/QUOTE]
Of course, but I conjecture this is the only instance where, for a given [$]n[/$], the [I]smallest[/I] prime of form [$]F_n'(a,b)=\frac{a^{2^n}+b^{2^n}}{2}[/$] has [$]a[/$] and [$]b[/$] both perfect squares. /JeppeSN 
[QUOTE=JeppeSN;483252]For both a and b odd, the first primes are (as others already found but did not post):
[CODE] [snip] (5^8+3^8)/2 [snip] (67^1024+57^1024)/2 [/CODE] [/QUOTE] Sheesh, I don't know HOW I messed that one up. I had posted [url=http://www.mersenneforum.org/showpost.php?p=482841&postcount=63]here[/url] (1^8 + 9^8)/2. But (3^8 + 5^8)/2 is correct. I also posted, for e = 2^11, n = (49^e + 75^e)/2. 
[QUOTE=JeppeSN;483260]Of course, but I conjecture this is the only instance where, for a given [$]n[/$], the [I]smallest[/I] prime of form [$]F_n'(a,b)=\frac{a^{2^n}+b^{2^n}}{2}[/$] has [$]a[/$] and [$]b[/$] both perfect squares. /JeppeSN[/QUOTE]
Not necessarily. It is the other way around  if the hit for the F[SUB]n+1[/SUB](a,b) happens very early, then it will decide the fate of the minimum in F[SUB]n[/SUB](a[SUP]2[/SUP],b[SUP]2[/SUP]). Here, in sequence [URL]https://oeis.org/A246119[/URL], the same happened three time already. 
[QUOTE=Batalov;483280]Not necessarily. It is the other way around  if the hit for the F[SUB]n+1[/SUB](a,b) happens very early, then it will decide the fate of the minimum in F[SUB]n[/SUB](a[SUP]2[/SUP],b[SUP]2[/SUP]).
Here, in sequence [URL]https://oeis.org/A246119[/URL], the same happened three time already.[/QUOTE] Squares are rare, so if either a or b is a square it is worthy of note. And, lo and behold, for m = 11 (though my results should be checked), one of the numbers (49, 75) is a square. So I wouldn't dismiss the possibility of another instance of [i]both[/i] being squares out of hand. Alas, the number of cases any of us will ever know of is, likely, quite small. 
[QUOTE=Dr Sardonicus;483283]Squares are rare, so if either a or b is a square it is worthy of note. And, lo and behold, for m = 11 (though my results should be checked), one of the numbers (49, 75) is a square. So I wouldn't dismiss the possibility of another instance of [i]both[/i] being squares out of hand.
Alas, the number of cases any of us will ever know of is, likely, quite small.[/QUOTE] They are more common, than the solutions to the next stage up... I think you mean minimal ones are of note. 
[QUOTE=Dr Sardonicus;483269]Sheesh, I don't know HOW I messed that one up. I had posted [url=http://www.mersenneforum.org/showpost.php?p=482841&postcount=63]here[/url] (1^8 + 9^8)/2. But (3^8 + 5^8)/2 is correct.
I also posted, for e = 2^11, n = (49^e + 75^e)/2.[/QUOTE] Apparently I forgot your post. Sorry. I agree on exponent 2^11. In the search for the next one, I find: [CODE](157^(2^12)+83^(2^12))/2 is 3PRP! (4.1038s+1.9638s) (157^(2^12)+111^(2^12))/2 is 3PRP! (4.0938s+3.3192s) (161^(2^12)+157^(2^12))/2 is 3PRP! (4.1053s+1.9778s)[/CODE] /JeppeSN 
[QUOTE=Batalov;483280]Not necessarily. It is the other way around  if the hit for the F[SUB]n+1[/SUB](a,b) happens very early, then it will decide the fate of the minimum in F[SUB]n[/SUB](a[SUP]2[/SUP],b[SUP]2[/SUP]).
Here, in sequence [URL]https://oeis.org/A246119[/URL], the same happened three time already.[/QUOTE] Yes, also interesting. I could be wrong, of course. With [I]two[/I] "free parameters" [$]a,b[/$] in [$]F_n'(a,b)[/$] is it likely to occur again for [$]n>11[/$]? I do not think so. The form in A246119, [$]k^{2^n} (k^{2^n}1)+1[/$], has only one parameter [$]k[/$]. I still think this square phenomenon should occur only finitely many times? /JeppeSN 
[QUOTE=JeppeSN;483297]Apparently I forgot your post. Sorry. I agree on exponent 2^11.
In the search for the next one, I find: [CODE](157^(2^12)+83^(2^12))/2 is 3PRP! (4.1038s+1.9638s) (157^(2^12)+111^(2^12))/2 is 3PRP! (4.0938s+3.3192s) (161^(2^12)+157^(2^12))/2 is 3PRP! (4.1053s+1.9778s)[/CODE] /JeppeSN[/QUOTE] [B]EDIT[/B] The below numbers are not primes. PFGW picks an "unlucky" base: And then: [CODE](3^(2^13)+1^(2^13))/2 is 3PRP! (0.6454s+0.4075s)[/CODE] EDIT: And again: [CODE](3^(2^14)+1^(2^14))/2 is 3PRP! (2.8285s+1.3930s)[/CODE] And: [CODE](9^(2^15)+1^(2^15))/2 is 3PRP! (57.8865s+21.0036s)[/CODE] So easy(?): [CODE](3^(2^16)+1^(2^16))/2 is 3PRP! (57.7878s+24.4478s)[/CODE] /JeppeSN 
Use base 2, obviously

[QUOTE=JeppeSN;483297]Apparently I forgot your post. Sorry. I agree on exponent 2^11.
In the search for the next one, I find: [CODE](157^(2^12)+83^(2^12))/2 is 3PRP! (4.1038s+1.9638s) (157^(2^12)+111^(2^12))/2 is 3PRP! (4.0938s+3.3192s) (161^(2^12)+157^(2^12))/2 is 3PRP! (4.1053s+1.9778s)[/CODE] /JeppeSN[/QUOTE] For exponent 2^13, we have: [CODE](107^(2^13)+69^(2^13))/2[/CODE] /JeppeSN 
[QUOTE=JeppeSN;483302][B]EDIT[/B] The below numbers are not primes. PFGW picks an "unlucky" base:
[code](3^(2^13)+1^(2^13))/2 is 3PRP! (0.6454s+0.4075s)[/CODE] [CODE](3^(2^14)+1^(2^14))/2 is 3PRP! (2.8285s+1.3930s)[/CODE] [CODE](9^(2^15)+1^(2^15))/2 is 3PRP! (57.8865s+21.0036s)[/CODE] [CODE](3^(2^16)+1^(2^16))/2 is 3PRP! (57.7878s+24.4478s)[/CODE] [/QUOTE] Hardly surprising. We have 3^2 == 1 (mod 2^3). Therefore, 3^(2^k) == 1 (mod 2^(k+2)) for every positive integer k. Now, let k be a positive integer, and N = (3^(2^k) + 1)/2. Then N  1 = (3^(2^k)  1)/2. By the above, we have 2^(k+1) divides N  1, so 3^(2^(k+1))  1 divides 3^(N  1)  1. Since 3^(2^(k+1))  1 = (3^(2^k)  1)*(3^(2^k) + 1), and (3^(2^k) + 1)/2 = N, we have N divides 3^(N1)  1. That is, N = 3^(2^k) + 1 is a base3 pseudoprime for every positive integer k. I haven't checked the RabinMiller criterion in this case, though. (The present instance is reminiscent of the fact that, if p is prime, 2^p  1 is a base2 pseudoprime, regardless of whether it's prime or composite.) 
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