2018-02-21, 03:02 | #1 |
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts |
Yes, my code from the thread has a slight typo, should say, y-1. But my point is primes with ⁴√-1 possible for all the bases may have 33+ forms divisible by them. Eliminating a lot of forms right off the start. For 17 these include:
Code:
2^(8x)*3^(16y+8)*5^(16z+8)*7^(16a+8)+1 2^(8x+4)*3^(16y)*5^(16z+8)*7^(16a+8)+1 2^(8x+4)*3^(16y+8)*5^(16z)*7^(16a+8)+1 2^(8x+4)*3^(16y+8)*5^(16z+8)*7^(16a)+1 2^(8x)*3^(16y)*5^(16z)*7^(16a+8)+1 2^(8x)*3^(16y)*5^(16z+8)*7^(16a)+1 2^(8x)*3^(16y+8)*5^(16z)*7^(16a)+1 2^(8x+4)*3^(16y)*5^(16z)*7^(16a)+1 forprime(x=1,10,forprime(y=x+1,100, if(y%4==1,for(z=1,y-1,if(lift(Mod(x,y)^z)==sqrt(Mod(-1,y)),print(x","y","z);next(2)))))) for all the sqrt of -1 cases( except PARI only knows one of them). And I think 1 mod 16 covers all 4th root of -1 cases. -1*-1*-1*1+1 \\ 4 ways if possible for all possibilities -1*1*1*1+1 \\ 4 ways sqrt(-1)*sqrt(-1)*-1*-1+1 \\ 6 ways; 12 ways if you use both sqrts possible sqrt(-1)*sqrt(-1)*1*1+1 \\ 6 ways; 12 ways if ... sqrt(-1)*sqrt(-1)*sqrt(-1)*(-sqrt(-1))+1 \\ 4 ways; 8 ways if ... sqrt(sqrt(-1))*sqrt(sqrt(-1))*sqrt(sqrt(-1))*sqrt(sqrt(-1))+1 \\ 1 way; maybe 4 if you use all 4 fourth roots. And maybe more. Last fiddled with by science_man_88 on 2018-02-23 at 23:12 Reason: Shortening thread |
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