Another failing 1-liner

[jnml]

? forprime(p=3,37,m=1<<p-1;n=1<<(p-2)-1;for(i=1,n,if(issquare(n),print(p);break);n+=m))
3
5
7
13
17
19
31
37
? ##
  ***   last result: cpu time 1min, 32,968 ms, real time 1min, 32,988 ms.
?

[science_man_88]

Originally Posted by jnml

? forprime(p=3,37,m=1<<p-1;n=1<<(p-2)-1;for(i=1,n,if(issquare(n),print(p);break);n+=m))
3
5
7
13
17
19
31
37
? ##
  ***   last result: cpu time 1min, 32,968 ms, real time 1min, 32,988 ms.
?

Is it supposedly to be 1<<(p-1) ??

[jnml]

Originally Posted by science_man_88

Is it supposedly to be 1<<(p-1) ??

AFAICT \(p-2\) is correct. It's about solving the old \(x^2+x+1=0 \pmod {m_p}\) again. (Completing the square produces \(n\) on the RHS.)

? forprime(p=3,31,print(p);m=1<<p-1;n=1<<(p-1)-1;for(i=1,n,if(issquare(n),print(p);break);n+=m))
3
5
7
11
13
17
19
23
29
31
? ##
*** last result: cpu time 4min, 35,941 ms, real time 4min, 36,209 ms.
?

[RMLabrador]

in the Hell, the Satan-machine exist, for coders (severe cases)

one rotation by right or left arm, the one step of cycle gone, all steps - you are free.
Be careful with 1-liners, I rotated it, I know...

[chalsall]

Originally Posted by RMLabrador

Be careful with 1-liners, I rotated it, I know...

I was always taught to simply manage the situation. As best as one can.

Read deeply. Use critical thought. Use the scientific method. Question everything; including oneself.

Understand the concept of deliberate noise meant to distract. Have thick skin. It's OK to be incorrect. It's also OK to be correct when everyone else thinks you are incorrect.

If I may... I will never be able to thank my teachers enough.

We are all students.

8^)

[jnml]

\(M_{p-2}^{M_{p-1}} \equiv 1 \pmod {M_p}\)

? m(p)=1<<p-1
%1 = (p)->1<<p-1
? forprime(p=3,4423,if(Mod(m(p-2),m(p))^m(p-1)==1,print(p)))
3
5
7
13
17
19
31
61
89
107
127
521
607
1279
2203
2281
3217
4253
4423
? ##
  ***   last result: cpu time 3,518 ms, real time 3,519 ms.
?

But that's Euler all over again, isn't it?