- Thread starter
- #1
Albert
Well-known member
- Jan 25, 2013
- 1,225
$n\in N$
$100\leq n\leq 1997$
$\dfrac{2^n+2}{n} $ also is an integer
please find n
$100\leq n\leq 1997$
$\dfrac{2^n+2}{n} $ also is an integer
please find n
I know! I know!$n\in N$
$100\leq n\leq 1997$
$\dfrac{2^n+2}{n} $ also is an integer
please find n
perl -e "use bigint; foreach $n (100..1997) { print $n if ((2**$n+2) % $n == 0); }"
946
Agh! I wanted to do that first on my HP-50g calculator. Alas, I am not familiar enough with programming it.Alternatively:
Code:perl -e "use bigint; foreach $n (100..1997) { print $n if ((2**$n+2) % $n == 0); }" 946
Same thing I am thinking. The problem can be rephrased asAlbert.Teng said:Can this be done without using computer program ?
I don't see an analytic way to derive this result, but there is a fairly easy way to verify that it is correct.Can this be done without using computer program ?
I mean using mathematical analysis only
As I said before, there is a great possibility that there are no other analytical method rather than brute-force.Albert.Teng said:but how to get the correct answer n=946 is a challenge
In the given interval, no. But in \(\displaystyle \mathbb{Z}\), infinitely many and perhaps density as much as \(\displaystyle \mathcal{O}(\log n)\).Albert.Teng said:Further more is there any other positive integers will also satisfy our needs ?
I have been watching this thread and am assuming Albert has an analytical solution (given this was posted in the Challenges subforum). I would be interested in the solution too, I've been playing around with the problem and am no closer to finding a solution.As I said before, there is a great possibility that there are no other analytical method rather than brute-force.
in fact we may follow the procedure to find n=946:$n\in N$
$100\leq n\leq 1997$
$\dfrac{2^n+2}{n} $ also is an integer
please find n
How does that follow? I don't understand.Albert.Teng said:(1 impossible)
take the last digit of $2^n$ into consideration 1 cannot appearYou can exclude the possibility that n is odd and hence, reducing your work a bit.
Originally Posted by Albert.Teng![]()
(1 impossible)
How does that follow? I don't understand.
mathbalarka said:You can exclude the possibility that n is odd and hence, reducing your work a bit.
Isn't it "2" instead of 3?$ 3^5$+1=33 is a multiple of 11
and $3^7$+1=129 is a multiple of 43 ,
Isn't it "2" instead of 3?
946. You didn't saw page 1 did you?What answers do the computer programme give?
So? Please elaborate.n/2 is also odd.
So we have to look only numbers of the form 2(mod 4).So our work is reduced further.So? Please elaborate.
The first 6 solutions the computer program gives, are:What answers do the computer programme give?
... Which is equivalent to 0 (mod 2) which is not true. Even numbers cannot satisfy the conditions in OP as I noted before. You can't say whether n/2 is odd or even since n/2 isn't really an integer. Hence, your claim that n/2 is odd is false.mathmaniac said:So we have to look only numbers of the form 2(mod 4)...
No.You said odd numbers can't satisfy the condition.... Which is equivalent to 0 (mod 2) which is not true. Even numbers cannot satisfy the conditions in OP as I noted before.
I can say.Here is my reasoning.You can't say whether n/2 is odd or even since n/2 isn't really an integer. Hence, your claim that n/2 is odd is false.
Hello mathbalarka.Same thing I am thinking. The problem can be rephrased as
\(\displaystyle 2^n = -2 \pmod{n}\)
Hence, n isn't prime. In fact, n is also not odd since if it's the case, n would be even and hence, creating a contradiction. See A006517, for example.
I don't there is any analytic way to prove it, the only thing would be brute-force search.
May I ?Hello mathbalarka.
I don't see how it follows that $n$ is not odd. Can you please elaborate.
[JUSTIFY]The argument does not follow. For instance, 14 is even, yet 7 (an odd number) divides it. What you are essentially saying is that no odd number divides an even number, which is clearly incorrect. This argument is only valid for powers of two, and unfortunately, $2^n + 2$ is never a power of two, except for the case $n = 1$.[/JUSTIFY]May I ?
2^n+2 is always even.If n is odd, \(\displaystyle \frac {2^n+2}{n}\) is not an integer and the question is to find n that satisfies \(\displaystyle \frac {2^n+2}{n}\) is an integer.
So n is not odd.
-mathmaniac