problem in number theory
Number Theory
Level
pending
Find all solutions in positive integers n, x, y, z, t of the equation n^x+n^y+n^z = n^t then find the value of n+y-2z+t+1 least value of answer is correct
1
4
2
3
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1 solution
We cannot have n = 1. If n = 2, then from (1) we get 1 +2^y-x + 2^z-x= 2^t-x and we cannot have y > x. Thus we have y = x, and 2+2^z-x = 2^t- x , which gives z-x = 1, hence t-x = 2. Thus, if n = 2, then we must have y = x, z = x+l, and t = x+2, while we easily check that for all positive integers x we have 2^x+2^x+2^x+1 = 2^x+2. Suppose, next, that n ≥ 3. In view of (1) we have 1 +n^y-x+n^z-x = n^t-x, and since n > 2, we must have y = x and z = x. Thus 3 = n^t-x, which implies n = 3 and t-x = 1. Therefore, if n > 2, we must have n = 3, x = y = z, t'= x+l. We easily check that for every positive integer x we have 3^x+3^x+3^x = 3^x+1
Thus, all solutions of equation (I) in positive integers n, x, y, z, t with z≥y≥x are n = 2, y = x, z = x+1, t = x+2, or n = 3, y = x, z = x, t = x+ 1, where x is an arbitrary positive integer.