2^x-41

enzocreti · 2019-01-21, 10:15

Are 8=x and 36=x the only integer values such that 2^x-41 is divisible by 215?

paulunderwood · 2019-01-21, 10:36

Quote:

Originally Posted by enzocreti

Are 8=x and 36=x the only integer values such that 2^x-41 is divisible by 215?

No. For example:

Code:

n=215;ph=eulerphi(n);e=8+ph*2^1000;r=Mod(2,n)^(e)-41;if(r==0,print(e))
  *** _^_: Warning: Mod(a,b)^n with n >> b : wasteful.
1800134460072929099193354082420803041743160083665296460505500652462189765889892685788573276410369041654359050501489286666314404079963087593245409032549360567061007386170470805550740533650478351835496302601711807797241766132922992949939811382940946569647961008251339084166952211717640896988650552235655176

For those values less than or equal to eulerphi(215):

Code:

n=215;ph=eulerphi(n);for(k=1,ph,r=Mod(2,n)^(k)-41;if(r==0,print(k)))
8
36
64
92
120
148

2019-01-21, 10:15	#1
enzocreti Mar 2018 530₁₀ Posts	2^x-41 Are 8=x and 36=x the only integer values such that 2^x-41 is divisible by 215?