(x+y)(x+z)(y+z)=2017^n
What is the minimum value n of such that there exist positive integers x,y and z satisfying the equation above?
x=y=z=1/2, n=0
Sorry, above is the only (non-integer) solution I could find.
To find the minimum value of n such that there exist positive integers x, y, and z satisfying the equation (x+y)(x+z)(y+z) = 2017^n, we need to analyze the given equation and understand its properties.
Since the equation involves the product of three factors, (x+y), (x+z), and (y+z), we can make a few observations:
1. The factors (x+y), (x+z), and (y+z) must all be divisors of 2017^n because the equation holds true. Additionally, they must be positive integers.
2. The factors (x+y), (x+z), and (y+z) are all greater than or equal to 2. This is because each factor represents a sum of at least two positive integers, and the smallest positive integers are 1 and 1, which add up to 2.
3. The factors (x+y), (x+z), and (y+z) are all less than or equal to 2017^n. This is because each factor is a sum of positive integers, and the largest possible sum would occur when each integer is equal to 1. In this case, the maximum value of each factor would be 2017.
Based on these observations, we can conclude that the minimum value of n can be determined by finding the largest power of 2017 that is a common divisor of all numbers that are 2 or greater and less than or equal to 2017.
Since 2017 is a prime number, the common divisor of all numbers less than or equal to 2017 is 2017^(1-1) = 2017^0 = 1. Therefore, the minimum value of n is 0.
This means that for any positive integers x, y, and z, the equation (x+y)(x+z)(y+z) = 2017^0 = 1 is satisfied.