 mersenneforum.org > Math Is this to prove/already known?
 Register FAQ Search Today's Posts Mark Forums Read 2007-06-25, 19:22 #1 MatWur-S530113   Apr 2007 Spessart/Germany 2×34 Posts Is this to prove/already known? Hello, I got this conjecture last night, but my math is not good enough to prove it... if and only if is divisible by then there is a non-negative Integer k with (= M(k) + 3) another conjecture: if n>1, n odd then is divisible by are these conjectures already known/proven? And if not, any idea how to prove them? I got them while looking in a self-created file with the (start of) factorisations of numbers of the form: , p prime (I made TF in this file up to 30 bit, if wanted I'll post it). Thanks in advance. mfg Matthias   2007-06-25, 21:12 #2 MatWur-S530113   Apr 2007 Spessart/Germany 2·34 Posts hmhmhmhm, something I made wrong. If then n is even, then n-1 is odd, then n-1 is not divisible by 2 as needed for . I made the division with a ShR-order, so the last bit was ignored. A corrected form of the first conjecture is: if and only if is divisible by then there is a non-negative Integer k with mfg Matthias   2007-06-26, 10:58   #3
maxal

Feb 2005

22×32×7 Posts Quote:
 Originally Posted by MatWur-S530113 if and only if is divisible by then there is a non-negative Integer k with
Consider two cases:

1) n is even then .
We have n=2m+2 and , implying that for some k. Therefore, n=2*2^k+2=2^(k+1)+2.

2) n is odd then .
We have n=2m+1 and , implying that and .   2007-06-27, 04:26   #4
MatWur-S530113

Apr 2007
Spessart/Germany

2×34 Posts Quote:
 Originally Posted by maxal Consider two cases: 1) n is even then . We have n=2m+2 and , implying that for some k. Therefore, n=2*2^k+2=2^(k+1)+2. 2) n is odd then . We have n=2m+1 and , implying that and .

1) if why does this imply that for some k? (Probably I miss an easy thought... )

2) if n is even I still don't see why n must have the form , if is divisible by (not only by ).

greetings,

Matthias   2007-06-27, 05:35   #5
maxal

Feb 2005

22·32·7 Posts Quote:
 Originally Posted by MatWur-S530113 Many thanks for your reply. But 2 things I don't understand: 1) if why does this imply that for some k? (Probably I miss an easy thought... )
Because means that m divides a power of 2, namely, . Therefore, for some k.

Quote:
 Originally Posted by MatWur-S530113 2) if n is even I still don't see why n must have the form , if is divisible by (not only by ).
If then , and I showed that the latter congruence implies n=2^(k+1)+2 for some k.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post George M Miscellaneous Math 5 2018-01-02 11:11 PawnProver44 Miscellaneous Math 40 2016-03-19 07:33 siegert81 Math 2 2014-11-19 10:24 Damian Math 31 2008-10-03 02:11 mdjvz Software 4 2003-09-28 17:13

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