Question about a mersennenumber property
I discovered a nice property about mersenne numbers.
[TEX](2^a+2^b)^{n}\equiv 2^a + 2^b\; (mod\; n)[/TEX] When [TEX] n=2^{p}1 [/TEX] is not prime then this is true only for the two trivial cases: 1.) [TEX]a = b[/TEX] 2.) If there exist an [TEX]x[/TEX] for : [TEX] 2^a+2^b \equiv 2^x[/TEX] the above congruence is also true. This is the case when choosing [TEX]a[/TEX] and [TEX]b[/TEX] , so that [TEX]ba \equiv p\;(mod\;n)[/TEX] .  For the other combinations of the variable [TEX]a[/TEX] and [TEX]b[/TEX] the above congruence is NEVER true. I do not really know if I am right with this. Up til now i haven't found an counterexample. :grin: My question is: Can this conjecture be true or is this just an example for the 'laws of small numbers' ? 
[QUOTE=sascha77;254247]I discovered a nice property about mersenne numbers.
[TEX](2^a+2^b)^{n}\equiv 2^a + 2^b\; (mod\; n)[/TEX] When [TEX] n=2^{p}1 [/TEX] is not prime then this is true only for the two trivial cases: 1.) [TEX]a = b[/TEX] 2.) If there exist an [TEX]x[/TEX] for : [TEX] 2^a+2^b \equiv 2^x[/TEX] the above congruence is also true. This is the case when choosing [TEX]a[/TEX] and [TEX]b[/TEX] , so that [TEX]ba \equiv p\;(mod\;n)[/TEX] .  For the other combinations of the variable [TEX]a[/TEX] and [TEX]b[/TEX] the above congruence is NEVER true. I do not really know if I am right with this. Up til now i haven't found an counterexample. :grin: My question is: Can this conjecture be true or is this just an example for the 'laws of small numbers' ?[/QUOTE] well if I did the math correct the only other time it could be true is a= b1 so can you come up with the answer from that ?, okay never mind messed up my math. 
[QUOTE=sascha77;254247]I discovered a nice property about mersenne numbers.
[TEX](2^a+2^b)^{n}\equiv 2^a + 2^b\; (mod\; n)[/TEX] When [TEX] n=2^{p}1 [/TEX] is not prime then this is true only for the two trivial cases: 1.) [TEX]a = b[/TEX] 2.) If there exist an [TEX]x[/TEX] for : [TEX] 2^a+2^b \equiv 2^x[/TEX] the above congruence is also true. This is the case when choosing [TEX]a[/TEX] and [TEX]b[/TEX] , so that [TEX]ba \equiv p\;(mod\;n)[/TEX] .  For the other combinations of the variable [TEX]a[/TEX] and [TEX]b[/TEX] the above congruence is NEVER true. I do not really know if I am right with this. Up til now i haven't found an counterexample. :grin: My question is: Can this conjecture be true or is this just an example for the 'laws of small numbers' ?[/QUOTE] This piece of stupidity is another example of what I say when I tell people that they should not go near number theory until they have mastered high school mathematics. Hint: [b]Binomial Theorem[/b]. Consider (x + y)^n mod n for arbitrary x,y \in N and n prime. 
well 2^x  2^(xz) = (2^z1)*2^(xz) since I redid the math. From the fact that (2^y)/(2^a) = 2^(ya) it's impossible to add up to a multiplier (2^z1) with only one other power of 2 other than 2^(xz) [COLOR="Red"]except[/COLOR] when z == 1.

Inb4miscmaththreads.

[QUOTE=science_man_88;254252]well 2^x  2^(xz) = (2^z1)*2^(xz) since I redid the math. From the fact that (2^y)/(2^a) = 2^(ya) it's impossible to add up to a multiplier (2^z1) with only one other power of 2 other than 2^(xz) [COLOR=Red]except[/COLOR] when z == 1.[/QUOTE]
I think you did really a mistake.:smile: [TEX]2^x  2^{(xz)} = 2^x  \frac{2^x}{2^z} = \frac{2^x*(2^z1)}{2^z}[/TEX] 
[QUOTE=sascha77;254256]I think you did really a mistake.:smile:
[TEX]2^x  2^{(xz)} = 2^x  \frac{2^x}{2^z} = \frac{2^x*(2^z1)}{2^z}[/TEX][/QUOTE] [TEX]\frac{2^x*(2^z1)}{2^z}[/TEX] according to your math you can reconvert it to 2^(xz)*(2^z1) which means unless you say the last one is a error my math holds up to yours. 
[QUOTE=R.D. Silverman;254251]This piece of stupidity is another example of what I say when I tell people
that they should not go near number theory until they have mastered high school mathematics. Hint: [B]Binomial Theorem[/B]. Consider (x + y)^n mod n for arbitrary x,y \in N and n prime.[/QUOTE] I think you did not read carefully what i wrote or my english grammar was too bad. Let me show you what I did and what i meant: I played with the Lemma : [TEX](Xa)^n \equiv\; (X^n a) \;(mod \;n)[/TEX] if [TEX]gcd(a,n) = 1[/TEX] and the congruent is correct then n is prime. And shure I know that I dont may put a number to X. X is a free variable. For example: [TEX](X1)^5 = 1 + 5X 10X^2 + 10X^35X^4 + X^5[/TEX] [TEX]1 + 5X 10X^2 + 10X^35X^4 + X^5 \;mod \;5\; = \;4 + X^5[/TEX] > 5 is prime But what I did was to play a little around with this Lemma. So I inserted the number 2 to the variables X and a. I know that for every [B]Mersenne[/B] [B]Prime[/B] [B]n[/B] (n=(2^p)1, n is prime) this is always true: [TEX](2^a+2^b)^n \equiv 2^a+2^b\;(mod\; n)[/TEX] (This is because the "Kleine Fermatsche Satz".) [TEX](2^a+2^b)^n \equiv 2^a+2^b\;(mod\; n)[/TEX] [TEX]\leftarrow \rightarrow\;\;(2^a+2^b)^{n1} \equiv 1\;(mod\; n)[/TEX] "Kleine Fermatsche Satz" : For all [TEX]a[/TEX] and all [TEX]n[/TEX] true: [TEX]a^{\varphi{(n)}} \equiv 1\; (mod \;n) [/TEX] [B]If n [/B][B]is prime then[/B] [TEX]\varphi{(n)}=n1[/TEX] ( [TEX]\varphi{}[/TEX] is the Eulers totient function ) Therefore if [TEX] n=2^p1[/TEX] is prime when for all [TEX]a[/TEX] and all [TEX]b[/TEX] the following congruation is always correct: [TEX](2^a+2^b)^n \equiv 2^a+2^b\;(mod\; n)[/TEX] ...And if you have : [TEX](x)^n \equiv x\; (mod \;n) [/TEX] you get this: [TEX]> x^{(n1)} \equiv 1 \;(mod \;n)[/TEX] But if n is not prime, when it can be that for some x this is also true. So what I did is that I looked for numbers x where the congruation [TEX]x^{(n1)} \equiv 1\; (mod \;n ) [/TEX] is [B]NEVER POSSIBLE[/B] when [B]n[/B] is [B]NOT PRIME[/B]. So I found a pattern. And this was my conjecture. Again: [TEX](2^a+2^b)^n \equiv 2^a+2^b \;(mod \;n) [/TEX] Let [TEX]n = 2^{p}1[/TEX] and [B]n[/B] is [B]not Prime[/B]. If now [TEX]a\not=b[/TEX] [B]and[/B] [TEX]ab\not\equiv \;p\;(mod\;n)[/TEX] then the above congruation is ALWAYS false. This means that for these [TEX]x = (2^a+2^b)[/TEX] this : [TEX]x^{(n1)}\equiv 1 \;(mod \;n)[/TEX] is [B]NEVER true [/B] when n=2^p1 is [B]not prime[/B] ! 
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Sascha,
Bob Silverman is an [B]expert[/B]. He likely did not misread what you wrote. If he suggests that you look into some area of education, do it. I think that his idea of High School math is through Math Analysis and Calculus I & II. He is often blunt, don't be scared by him though. I have not looked into your suggestion. My math skills have severely atrophied since I left school. [SIZE="1"][COLOR="Silver"]semirandom image attached.[/COLOR][/SIZE] 
[QUOTE=sascha77;254261]
[TEX](2^a+2^b)^n \equiv 2^a+2^b \;(mod \;n) [/TEX] Let [TEX]n = 2^{p}1[/TEX] and [B]n[/B] is [B]not Prime[/B]. If now [TEX]a\not=b[/TEX] [B]and[/B] [TEX]ab\not\equiv \;p\;(mod\;n)[/TEX] then the above congruation is ALWAYS false. This means that for these [TEX]x = (2^a+2^b)[/TEX] this : [TEX]x^{(n1)}\equiv 1 \;(mod \;n)[/TEX] is [B]NEVER true [/B] when n=2^p1 is [B]not prime[/B] ![/QUOTE] Well, (2[SUP]11[/SUP]+2[SUP]1[/SUP])[SUP]2046[/SUP] [TEX]\equiv[/TEX] 1 (mod 2047). ab=10 [TEX]\not\equiv[/TEX] 11 (mod 2047) 
[QUOTE=sascha77;254261]
This means that for these [TEX]x = (2^a+2^b)[/TEX] this : [TEX]x^{(n1)}\equiv 1 \;(mod \;n)[/TEX] is [B]NEVER true [/B] when n=2^p1 is [B]not prime[/B] ![/QUOTE] You just [b]changed[/b] your original statement. Your original statement was that x^n = x is not true when n = 2^p1 is composite. Here x = 2^a + 2^b for a!= b This is different from x^{n1} = 1 is not true when n = 2^p1 is composite. 
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