answer:To determine whether the distribution functions of xi zeta and eta zeta are the same, even though the distribution functions of xi and eta are the same, we will consider a counterexample. 1. **Set Up the Counterexample:** - Let zeta := xi. This means that zeta is identically equal to xi. 2. **Compute xi zeta:** - Given zeta := xi, we have: [ xi zeta = xi cdot xi = xi^2 ] Since xi is a random variable, xi^2 is also a random variable. However, a notable property is that if xi takes values of 1 and -1 with equal probability, then: [ xi^2 equiv 1 ] This is because both 1^2 = 1 and (-1)^2 = 1. 3. **Compute eta zeta:** - Since zeta = xi and xi and eta have the same distribution, consider eta also taking values of 1 and -1 with equal probability. - Then: [ eta zeta = eta xi ] Since both eta and xi are independent from each other but identically distributed, eta xi can be calculated. If xi = 1 or xi = -1: [ eta xi = eta ] If eta = 1 and xi = -1, then eta xi = -1. Therefore, eta xi can take both values 1 and -1. 4. **Compare Distribution Functions:** - The random variable xi^2 always equals 1. - The random variable eta xi can take both values 1 and -1, with equal probability. Thus, the distribution functions of xi zeta and eta zeta are not the same. # Conclusion Hence, the distribution functions of xi zeta and eta zeta do not coincide, as demonstrated by the example. boxed{text{No}}

answer:To find the number of unique ways to sum prime numbers to get 8, we need to consider all possible combinations of prime numbers that add up to 8, with the condition that each prime number in the sum is less than or equal to the next prime number in the sum (i.e., a ≤ b ≤ c ≤ d ...). The prime numbers less than 8 are 2, 3, 5, and 7. We need to find combinations of these prime numbers that add up to 8. Let's consider the possible combinations: 1. Using the prime number 2: - 2 + 2 + 2 + 2 = 8 (four 2s) - There are no other combinations using only the prime number 2 that add up to 8. 2. Using the prime number 3: - There are no combinations of 3s that add up to 8 because 3 + 3 = 6 and 3 + 3 + 3 = 9, which is too high. 3. Using the prime number 5: - There are no combinations of 5s that add up to 8 because 5 is already greater than 3, and we cannot use 2 + 5 because that would only sum to 7. 4. Using the prime number 7: - There are no combinations of 7s that add up to 8 because 7 is already greater than 8. 5. Using a combination of different prime numbers: - 2 + 2 + 2 + 2 = 8 (already considered) - 2 + 3 + 3 = 8 (one 2 and two 3s) - There are no other combinations of different prime numbers that add up to 8. From the above considerations, we have the following unique ways to sum prime numbers to get 8: 1. 2 + 2 + 2 + 2 2. 2 + 3 + 3 Therefore, there are 2 unique ways to sum prime numbers to get 8, so f(8) = boxed{2} .

answer:To find the percentage increase, you can use the following formula: Percentage Increase = [(Final Number - Initial Number) / Initial Number] * 100 In this case, the initial number is 350 and the final number is 525. Percentage Increase = [(525 - 350) / 350] * 100 Percentage Increase = [175 / 350] * 100 Percentage Increase = 0.5 * 100 Percentage Increase = 50% So, the percentage increase is boxed{50%} .

answer:Let ( p ) be an odd prime number and ( n ) be a positive integer. We aim to prove that the number ( p n^{2} ) has at most one positive divisor ( d ) such that ( d+n^{2} ) is a perfect square. **First Solution:** 1. **Prime Factorization and Case Analysis:** According to the Fundamental Theorem of Arithmetic, ( d ) must be composed of prime factors from the prime factorization of ( p n^{2} ), and the exponents of these factors in ( d ) must not exceed those in ( p n^{2} ). We have two cases: - If ( p ) does not appear in the factorization of ( d ), then ( d ) divides ( n^{2} ). - If ( p ) appears in the factorization of ( d ), then ( d = p d^{prime} ), where ( d^{prime} ) divides ( n^{2} ). 2. **Existence of an Integer ( m ):** By the given condition, there exists a positive integer ( m ) such that in the first case: [ d + n^{2} = m^{2} ] and in the second case: [ p d^{prime} + n^{2} = m^{2} ] 3. **Prime Divisors Analysis:** Consider an arbitrary prime divisor ( q ) of ( d ) or ( d^{prime} ). Then, since ( q ) also divides ( n^{2} ) and ( m^{2} ), by the essential uniqueness of prime factorization, the bases ( n ) and ( m ) must also be divisible by ( q ). Thus ( n^{2} ) and ( m^{2} ) must also be divisible by ( q^{2} ). 4. **Simplification:** Consequently, ( d ) or ( d^{prime} ) must be divisible by ( q^{2} ), except in the second case when ( q = p ). Simplifying by ( q^{2} ), we obtain: [ d^{*} + n^{prime 2} = m^{prime 2}, qquad text{or} qquad p d^{*} + n^{prime 2} = m^{prime 2} ] where ( d^{*} ) divides ( n^{prime 2} ). 5. **Iterative Simplification:** Repeating the process, we find that there exist positive integers ( n_{1} ) and ( m_{1} ) such that: [ 1 + n_{1}^{2} = m_{1}^{2} quad text{or} quad p + n_{1}^{2} = m_{1}^{2} ] 6. **Square Difference Analysis:** The first equation is impossible because the difference between two positive squares must be at least 3. Therefore, for the second equation: [ p = (m_{1} + n_{1})(m_{1} - n_{1}) ] Given that ( p ) is prime, we have: [ m_{1} + n_{1} = p, quad m_{1} - n_{1} = 1 quad Rightarrow quad 2 n_{1} = p - 1 ] 7. **Conclusion on ( n ) and ( d ):** This implies: [ n = frac{1}{2}(p-1) n_{2} quad text{and} quad d = p n_{2}^{2} ] 8. **Verification:** Substituting these values: [ d + n^{2} = left[ p + left( frac{1}{2}(p-1) right)^{2} right] n_{2}^{2} = left[ frac{1}{2}(p+1) right]^{2} n_{2}^{2} = left[ frac{1}{2}(p+1) n_{2} right]^{2} ] which is indeed a perfect square. Hence, there is at most one such divisor ( d ), and it exists if and only if ( n ) is divisible by (frac{1}{2}(p-1)). (blacksquare)