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October 11th, 2017, 07:11 PM   #1
GGQ
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Help for the hard prove question

If n is a positive integer and n is even,

prove: (2^(n!)-1) is divisible by (n^2-1).

This question confuse me some days.

Please help me or give me some hint. Thank you very much.

Last edited by skipjack; October 12th, 2017 at 02:57 AM.
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October 11th, 2017, 08:27 PM   #2
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Quote:
Originally Posted by GGQ View Post
If n is a positive integer and n is even,

prove: (2^(n!)-1) is divisible by (n^2-1).

This question confuse me some days.

Please help me or give me some hint. Thank you very much.
Let n = 3. Then $\displaystyle \frac{2^{3!} - 1}{3^2 - 1} = \frac{2^6 - 1}{8}$ $\displaystyle = \frac{63}{8}$
So it doesn't work for n = 3.

-Dan

Last edited by skipjack; October 12th, 2017 at 02:58 AM.
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October 11th, 2017, 08:44 PM   #3
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question is " n is even "
so the example that you take n=3 is illegal.
thank you.
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October 12th, 2017, 12:35 PM   #4
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Originally Posted by GGQ View Post
question is " n is even "
so the example that you take n=3 is illegal.
thank you.
Sorry. Yes, I missed that.

-Dan
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October 12th, 2017, 01:29 PM   #5
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Since n is an even positive integer, we can take n= 2k for k any positive integer. Since we have a statement that depends on the positive integer k, this pretty much cries out for "proof by induction" on k.

Base case: k= 1. In that case n= 2, n!= 2 so the statement becomes "2^2- 1= 4- 1= 3 is divisible by 2^2- 1= 4- 1= 3. 3 is clearly divisible by 3.

Now, suppose that 2^((2k)!-1) is divisible by (2k)^2- 1. That is: 2^((2k)!- 1)= p((2k)^2- 1) for some integer, p. Then 2^((2(k+1))!- 1)= 2^(2(k+1)(2k+1)(2k!)- 1)= 2^(2(k+1)(2k+1)(k!-1)+ k+1-1)= 2^k2^((k+1)(k!-1)= 2^k(2^(k!+ 1))^(k+1)= 2^k(p(k^2- 1))^{k+ 1). That last clearly has k^2- 1 as a factor so is divisible by k^2- 1.
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Last edited by Country Boy; October 12th, 2017 at 01:33 PM.
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October 12th, 2017, 06:58 PM   #6
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Don't you need it to be divisible by $(2k)^2-1=4k^2-1$?
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October 12th, 2017, 11:09 PM   #7
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Quote:
Originally Posted by Country Boy View Post
Since n is an even positive integer, we can take n= 2k for k any positive integer. Since we have a statement that depends on the positive integer k, this pretty much cries out for "proof by induction" on k.

Base case: k= 1. In that case n= 2, n!= 2 so the statement becomes "2^2- 1= 4- 1= 3 is divisible by 2^2- 1= 4- 1= 3. 3 is clearly divisible by 3.

Now, suppose that 2^((2k)!-1) is divisible by (2k)^2- 1. That is: 2^((2k)!- 1)= p((2k)^2- 1) for some integer, p. Then 2^((2(k+1))!- 1)= 2^(2(k+1)(2k+1)(2k!)- 1)= 2^(2(k+1)(2k+1)(k!-1)+ k+1-1)= 2^k2^((k+1)(k!-1)= 2^k(2^(k!+ 1))^(k+1)= 2^k(p(k^2- 1))^{k+ 1). That last clearly has k^2- 1 as a factor so is divisible by k^2- 1.

This part " 2^((2k)!-1) is divisible by (2k)^2- 1. " is wrong.
Correct is ( 2^(2k)!) -1 is divisible by (2k)^2- 1
thank you.
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