answer:Solution: (I) From the given information, we have: f(x)=sin omega xcos omega x- sqrt{3}cos^2omega x+ frac{sqrt{3}}{2} = frac{1}{2}sin 2omega x- frac{sqrt{3}}{2}cos 2omega x =sin (2omega x- frac{pi}{3}), Since the two adjacent axes of symmetry are frac{pi}{2}, the period T=pi, thus omega=1, Therefore, f(x)=sin (2x- frac{pi}{3}), From 2x- frac{pi}{3}= frac{pi}{2}+2kpi (kinmathbb{Z}), we get x= frac{5}{12}pi+kpi (kinmathbb{Z}), which is the equation of the axis of symmetry for the function y=f(x). (II) From (I), we know that the axis of symmetry for f(x) is x= frac{5}{12}pi+kpi (kinmathbb{Z}), Therefore, when xin(0,pi), the axis of symmetry is L: x= frac{5}{12}pi, Since the zeros of the function y=f(x)- frac{1}{3} in (0,pi) are x_{1} and x_{2}, which are symmetric about L, Therefore, x_{1}+x_{2}= frac{5}{6}pi, Thus, x_{1}= frac{5}{6}pi-x_{2}, Therefore, cos (x_{1}-x_{2})=cos ( frac{5}{6}pi-2x_{2}), =sin (2x_{2}- frac{pi}{3}), Since f(x_{2})= frac{1}{3}, Therefore, cos (x_{1}-x_{2})= frac{1}{3}. Thus, the final answers are: (1) The equation of the axis of symmetry for the function y=f(x) is x= frac{5}{12}pi+kpi (kinmathbb{Z}). (2) The value of cos (x_{1}-x_{2}) is boxed{frac{1}{3}}.

answer:1. Let ( n ) be a natural number, and let ( d ) be a divisor of ( 2n^2 ). 2. Suppose ( 2n^2 = kd ) where ( k ) is a positive integer since ( d ) is a divisor of ( 2n^2 ). 3. We need to show that ( n^2 + d ) is not a perfect square. 4. Assume, for the sake of contradiction, that ( n^2 + d = x^2 ) for some integer ( x ). 5. Then, we can write: [ x^2 = n^2 + d = n^2 + frac{2n^2}{k} = n^2 left(1 + frac{2}{k} right) ] 6. This implies: [ x^2 = n^2 left(1 + frac{2}{k} right) ] By expanding, we get: [ k^2 x^2 = n^2 (k^2 + 2k) ] 7. Observe that the left side, ( k^2 x^2 ), is a perfect square. 8. Now, for the right side ( n^2(k^2 + 2k) ) to be a perfect square, ( k^2 + 2k ) must also be a perfect square. 9. Consider the inequality: [ k^2 < k^2 + 2k < (k+1)^2 ] This shows that ( k^2 + 2k ) is strictly between two consecutive perfect squares, which means it cannot be a perfect square. 10. Therefore, our assumption that ( n^2 + d ) is a perfect square leads to a contradiction. Conclusion: Hence, ( n^2 + d ) cannot be a perfect square. [ boxed{} ]

answer:# Problem: 设 a_{1}, a_{2}, cdots, a_{2008} 为 2008 个整数，且 1 leqslant alpha_{i} leqslant 9 (i=1,2, cdots, 2008). 如果存在某个 k in {1,2, cdots, 2008}，使得 2008 位数 overline{alpha_{k} alpha_{k+1} cdots alpha_{2008} alpha_{1} cdots alpha_{k-1}} 被 101 整除，试证明: 对所有 i in {1,2, cdots, 2008}, 2008 位数 overline{alpha_{i} alpha_{i+1} cdots alpha_{2008} alpha_{1} cdots alpha_{i-1}} 均能被 101 整除。 1. 根据已知条件，设 k=1，即 2008 位数 alpha_{1} alpha_{2} cdots alpha_{2008} 能被 101 整除，我们只需证明 2008 位数 overline{alpha_{2} alpha_{3} cdots alpha_{2008} alpha_{1}} 也能被 101 整除。 2. 设 [ A = overline{alpha_{1} alpha_{2} cdots alpha_{2008}} = 10^{2007} alpha_{1} + 10^{2006} alpha_{2} + cdots + 10^1 alpha_{2007} + 10^0 alpha_{2008} ] 3. 则 [ B = overline{alpha_{2} alpha_{3} cdots alpha_{2008} alpha_{1}} = 10^{2007} alpha_{2} + 10^{2006} alpha_{3} + cdots + 10^1 alpha_{2008} + 10^0 alpha_{1} ] 4. 可以观察到 10A - B 的形式如下: [ 10A - B = 10 left(10^{2007} alpha_{1} + 10^{2006} alpha_{2} + cdots + 10^1 alpha_{2007} + 10^0 alpha_{2008} right) - left(10^{2007} alpha_{2} + 10^{2006} alpha_{3} + cdots + 10^1 alpha_{2008} + 10^0 alpha_{1} right) ] 5. 简化上面的表达式: [ 10A - B = 10^{2008} alpha_{1} - alpha_{1} = (10^{2008} - 1) alpha_{1} ] 6. 10^{2008} - 1 可以进一步分解成: [ 10^{2008} - 1 = (10^{4})^{502} - 1 = (10000)^{502} - 1 = [(9999 + 1)^{502} - 1] ] 7. 使用二项式定理展开: [ (9999 + 1)^{502} = 9999^{502} + binom{502}{1} 9999^{501} cdot 1 + cdots + 1^{502} ] 因为 9999 是 10^4 - 1，即 9999 = -2 (mod 101)，通过计算可以得出 (9999 + 1)^{502} - 1 默认是101的倍数。 ] 8. 因此，10A - B 是有 101 整除。 [ 10A - B equiv 0 pmod{101} ] 9. 因为 101 mid A 和 101 mid 9999，所以 101 mid B。即 B = overline{alpha_{2} alpha_{3} cdots alpha_{2008} alpha_{1}} 能被 101 整除。 ] 10. 使用上述方法依次类推出 i in {1, 2, cdots, 2008} 的情形，通过递推可以得到 2008 位数 overline{alpha_{i} alpha_{i+1} cdots alpha_{2008} alpha_{1} cdots alpha_{i-1}} 均能被 101 整除。 ] # Conclusion: [ boxed{} ]

answer:1. **Identify the total number of days and total cost**: Bob takes the pills for three weeks, which is 21 days. The total cost for these three weeks is 966. 2. **Calculate daily expenditure**: [ text{Daily cost} = frac{text{Total cost}}{text{Number of days}} = frac{966}{21} = 46 text{ dollars} ] 3. **Set up the equation for daily costs**: Let the cost of a blue pill be y dollars and the cost of a red pill be y-4 dollars. Since Bob takes one blue pill and one red pill each day, the combined daily cost is: [ y + (y - 4) = 2y - 4 ] 4. **Solve for y**: Since the daily cost is 46 dollars, we set up the equation: [ 2y - 4 = 46 ] Solving for y, we add 4 to both sides: [ 2y - 4 + 4 = 46 + 4 implies 2y = 50 ] Then, divide by 2: [ y = frac{50}{2} = 25 ] 5. **Conclusion**: The cost of one blue pill is 25 text{dollars}. The final answer is The correct choice is **boxed{textbf{(D) }textdollar 25}**.