Question

如何证明2^n-1不是完全平方数也不是完全立方数?

Answer 1

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It is easy
you can give a counter example
then you statement will be proved. Say 2^n-1 is a plete cubic or square of a number. Take n=2 2^2-1=3
where 3 is not a plete cubic or square of a number. There is contradiction. Then 2^n-1 is not a plete 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 补充： 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 补充： 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 补充： 2^k-1 = c where c is either a plete 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 补充： Then you will find it is difficult to prove it true or not. 2008-02-01 16:56:27 补充： So
we take a counter example is the most easiest way to disprove any statements. 2008-02-01 16:57:51 补充： 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.
心算证明：任何两个偶数相乘 ≡ 0 (mod 4)。〔广泛定义偶数为 2n 形式，n 为整数〕 任何两个奇数相乘 ≡ 1 (mod 4)。 2009-10-23 09:54:27 补充： 这一题用 mod 几行就证完。 以下之事实显然易见〔由前面意见之心算证明〕： 【事实】对任何整数 x，x^2 及 x^3 ≡ 0 or 1 (mod 4)。 首先 n = 1 时，2^n - 1 = 1 的确是平方数及立方数。所以，题目需假设 n > 1 才行。 接着观察对 n > 1
(2^n) - 1 ≡ 3 (mod 4)。 由【事实】，即得当 n > 1 时， (n^2)- 1 不可能为平方数或立方数。　
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不是完全平方数也不是完全立方数. 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不是完全平方数. 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不是完全立方数. 1)和2) 证明
当n>1时
2^n-1不是完全平方数也不是完全立方数.
这里要证明的是对所有大于1的自然数n
2^n-1 都不是平方或立方
因此只给出一个反例是不行的!. (当 n=1
2^n-1=1
既是平方也是立方) 证明如下: 若2^n-1 是平方
因为2^n-1是单数
所以它必定等于(2k+1)^2
其中k是某一自然数. 于是 2^n -1 = (2k+1)^2 = 4k^2+4k+1 2^n = 4k(k+1) + 2 注意左边是4的倍数
但右边不是4的倍数
因此矛盾. 所以当n大于1
2^n-1不可以是平方. 若2^n-1是立方
则它要等于(2k+1)^3. 于是 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))是单数
而且不等于1. (因为n大于1
2k+1不能等于1) 所以2^n有一个不等于1的单数的因数
这是不可能的. 因此当n大于1
2^n-1不可以是立方 2008-02-01 21:29:28 补充： 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 补充： 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 wering at all. 2008-02-01 21:33:28 补充： Obviously we understand the question differently
but I think you interpret the question too simply. 2008-02-01 21:34:33 补充： 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 补充： 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 wer it at all. 2008-02-01 21:37:53 补充： Yet I have proved it !!! 2008-02-01 21:39:46 补充： Last mend (at least meanwhile
unless you have further supplement):We are proving a statement in this question
not just disproving something! Think more carefully.