如何證明2^n-1不是完全平方數也不是完全立方數?

It is easy, you can give a counter example, then you statement will be proved.
Say 2^n-1 is a complete cubic or square of a number.
Take n=2
2^2-1=3, where 3 is not a complete cubic or square of a number. There is contradiction.
Then 2^n-1 is not a complete cubic or square of a number.
This is called prove by contradiction, assume it is correct and use a counter example to disprove it.

2008-02-01 16:51:36 補充：
This is the general methol to disprove a statement. Any counter example can disprove it. If anf only if all of the assumption was valid to prove the statement is true. In a easiler way to say, we need all crroect to prove a statement true, while 1 counter example can disprove it.

2008-02-01 16:53:01 補充：
So I take n=2 as example, you can take n=1 and start your prove like by MI, sayn=1, 2^n-1 = 1, then assume iti is true for a number k.

2008-02-01 16:54:59 補充：
2^k-1 = c where c is either a complete cubic or square of a number.Then prove for k 12^(k 1)-1=dLHS=2 (2^k-1) 1=2c 1, so we need to prove 2c 1 is a cubic or square of a number or not.

2008-02-01 16:55:29 補充：
Then you will find it is difficult to prove it true or not.

2008-02-01 16:56:27 補充：
So, we take a counter example is the most easiest way to disprove any statements.

2008-02-01 16:57:51 補充：
This is prove by contradiction. 1 COUNTER EXAMPLE IS ENOUGH. You need to get all correct only in the case of proving a statement true.

心算證明：任何兩個偶數相乘 ≡ 0 (mod 4)。〔廣泛定義偶數為 2n 形式，n 為整數〕

任何兩個奇數相乘 ≡ 1 (mod 4)。

2009-10-23 09:54:27 補充：
這一題用 mod 幾行就證完。
以下之事實顯然易見〔由前面意見之心算證明〕：

【事實】對任何整數 x，x^2 及 x^3 ≡ 0 or 1 (mod 4)。

首先 n = 1 時，2^n - 1 = 1 的確是平方數及立方數。所以，題目需假設 n > 1 才行。

接著觀察對 n > 1,
(2^n) - 1 ≡ 3 (mod 4)。

由【事實】，即得當 n > 1 時， (n^2)- 1 不可能為平方數或立方數。　

Excellent ~

当n=1时, 2^n-1=1, 1^2和1^3都是1. 所以问题应该是n>1, 2^n-1不是完全平方數也不是完全立方數.

证明如下: (用反证法)
1) 2^n-1 为奇数, 假设存在某自然数k, 使得 2^(n+1)-1=(2k+1)^2 成立, 则
2*2^n-1=(2k+1)^2
2(2^n-1) + 1 = (2k+1)^2
2(2^n-1) = (2k+1)^2 - 1
2(2^n-1) = (2k+1-1)(2k+1+1)
2(2^n-1) = 2k*2(k+1)
2^n-1 = 2k(k+1)
等式左边为奇数,右边为偶数, 矛盾. 故 2^n-1不是完全平方數.
2) 2^n-1 为奇数, 假设存在奇数2k-1>1 , 使得且 2^n - 1=(2k-1)^3 成立, 则
2^n = (2k-1)^3 + 1
2^n = (2k-1 + 1)( (2k-1)^2 - (2k-1) + 1 )
= 2k( 4k^2 - 4k +1 -2k +1 +1)
=2k(4k^2-6k+3)
=2k( 2(k-1)(2k-1)+1)
因为, 2^n 的奇数因数只有1, 而2(k-1)(2k-1)+1为奇数,且为2^n的因数, 所以
2(k-1)(2k-1)+1 = 1
2(k-1)(2k-1)=0
k为自然数, 所以k=1,
这与 2k-1>1 即 k>1 矛盾. 故2^n-1不是完全立方數.

1)和2) 证明,当n>1时, 2^n-1不是完全平方數也不是完全立方數.

這裡要證明的是對所有大於1的自然數n, 2^n-1 都不是平方或立方, 因此只給出一個反例是不行的!. (當 n=1, 2^n-1=1, 既是平方也是立方)

證明如下:

若2^n-1 是平方, 因為2^n-1是單數, 所以它必定等於(2k+1)^2, 其中k是某一自然數. 於是

2^n -1 = (2k+1)^2 = 4k^2+4k+1
2^n = 4k(k+1) + 2

注意左邊是4的倍數, 但右邊不是4的倍數, 因此矛盾. 所以當n大於1, 2^n-1不可以是平方.

若2^n-1是立方, 則它要等於(2k+1)^3. 於是

2^n - 1 = (2k+1)^3
2^n = (2k+1)^3 + 1 = ((2k+1)+1) ((2k+1)^2 - (2k+1) +1))

注意((2k+1)^2 - (2k+1) +1))是單數, 而且不等於1. (因為n大於1, 2k+1不能等於1)
所以2^n有一個不等於1的單數的因數, 這是不可能的. 因此當n大於1, 2^n-1不可以是立方

2008-02-01 21:29:28 補充：
giving one counter example can only disprove the statement2^n-1 is a square or a cube for all n.In other words, it only proved 2^n-1 is not a square nor a cube for SOME n.

2008-02-01 21:30:52 補充：
but the problem here is obviously to prove 2^n-1 is not a square nor a cube for ALL n. Otherwise it will not be a question worth answering at all.

2008-02-01 21:33:28 補充：
Obviously we understand the question differently, but I think you interpret the question too simply.

2008-02-01 21:34:33 補充：
By the way, to prove by contradiction does not always mean giving one counter example is enough. It depends on the type of the question.

2008-02-01 21:37:17 補充：
You are right that it is not so easy to prove it is not a cube, and that is the whole point of the question. If it is so easy, I won't answer it at all.

2008-02-01 21:37:53 補充：
Yet I have proved it !!!

2008-02-01 21:39:46 補充：
Last commend (at least meanwhile, unless you have further supplement):We are proving a statement in this question, not just disproving something! Think more carefully.