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T1-1. Propose that \(m > n \geqslant 0\), Prove that \( (2{2n} + 1) \mid (2{2m} - 1) \) .
Since we have:\[x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1})\]
Therefore, we can rewrite:\[
2{2m} - 1 = (2{2{n+1}}){2{m-n-1}} - 1{2{m-n-1}} = (2{2{n+1}} - 1)(2{2(2^{m-n-1}) - 1} + \cdots + 1)\]
Thus, we have \( 2{2{n+1}} - 1 \mid 2{2m} - 1\). Since \(2{2{n+1}} - 1 = (2{2n})^2 - 1 = (2{2n} - 1)(2{2n} + 1)\), Therefore \( (2{2n} + 1) \mid (2{2m} - 1) \).
T1-2. Suppose \(k \geqslant 1 \) is an odd integer, Prove that \(1^k + 2^k +3^k +\cdots +n^k\) can not be divisible by \(n+2\).
When \(n=1\) it must be true. When \(n \geqslant 2\), we can rewrite the summation(A) as follows:\[2A = 2 + (2^k + n^k) + (3^k + (n -1)^k) + \cdots + (n/2)^k + (n/2 + 1)^k\]
Since k is odd integer, for each \(i \geqslant 2\), we have:\[i^k + (n - i + 2)^k = (n + 2)(i^{k-1} + i^{k-2}(n-i+2) + \cdots + (n - i +2)^{k-1})\]Therefore, 2A divided by \( n + 2\) leaces a reminder of 2. Thus, \(A\) can not be divisible by \(n + 2\).
T1-3. Suppose a,m,n are all integers, \(a \geqslant 2\), show that:\(a^m - 1 \mid a^n - 1\) is a sufficient requarement of \(m\mid n.\)
If \(m \mid n\), then \(n=km\), \[ a^n - 1 = a^{km} - 1^k = (a^m - 1)(a^{(k-1)m} + a^{(k-2)m} + \cdots + a^m + 1)\]
Therfore \(a^m - 1 \mid a^n - 1\).
If \(a^m - 1 \mid a^n - 1\), then we can write:\[ n =km + r, 0\leqslant r < m\]
Then we have:\[a^n - 1 = a^{mk + r} - 1 = a^r (a^{mk} - 1) + (a^r - 1) \]
Since \(a^m - 1 \mid a^n - 1\), we have \(a^m - 1 \mid a^r - 1\), while \(0\leqslant r < m\) , thus \(r = 0\). therfore \(m\mid n\).
T1-4. \(n > 0\), Show that: there is an integer a, which can make \(a^a +1, a{aa} + 1,\cdots\) can all be divisible by n.
Suppose \(a = 2k + 1\), then \(a^a = (2k + 1)^{(2k+ 1)}\). Since the multiplication of the odd number is odd, thus \(a^a, a{aa},\cdots\) are all odd.
Since \[a^a + 1,a{a{a}} + 1,c\dots\] are all have the factor of \((a+1)\), thus we can let \(a+1 = 2n, a=2n-1\).
T1-5.\(n\geqslant 2\),Show that there are n mutually different positive integer number, with the multiplication of these n number divisible the summary of arbitary two number.
First we chose n different number \(n_1, n_2, c=\cdots, n_n\), and then let every item multiplies a factor K. Then we want each two item \(Kn_i + Kn_j\) to be divisible by \(K^n\Pi n_i\). Then we can take K = \(\Pi (n_i+n_j)\).
T1-6 Suppose \(2^{k_1} + 2^{k_2} + \cdots + 2^{k_n} \) is divisible by \(2^m -1\), show that \(n \geqslant m\).
Suppose \(n < m\)
We consider the worst condition with n different \(k_i\) since the same \(k_i\) will make us get \(2^{k_i+1}\),which makes \(n\) smaller. And this condition will let us get the smallest \(n\).
Then we can rewrite the power as follows:\[2^{k_i} = 2^{mq_i+r_i},0\leqslant r_i < m\]
Since \(2^m \equiv 1 \mod (2^m - 1)\), we have:\[2^{mq_i+r_i} \equiv 2^{r_i} \mod (2^m - 1)\]\[2^{k_1} + 2^{k_2} + \cdots + 2^{k_n} \equiv 2^{r_1} + 2^{r_2} + \cdots + 2^{r_n} \mod (2^m - 1)\]
Since \[2^{r_1} + 2^{r_2} + \cdots + 2^{r_n} \leqslant 2^1 + \cdots + 2^{m-1} = 2^m - 2 < 2^M -1\]