Let N be the set of positive integers. Find all functions f:N->N that satisfy the equation
fabc-a(abc) + fabc-b(abc) + fabc-c(abc) = a + b + c
for all a, b, c ≥ 2.
(Here f1(n) = f(n) and fk(n) = f(fk-1(n)) for every integer k greater than 1)
(Also note: abc is the product a·b·c and not the concatenation 100a+10b+c)
We can discover constraints on f(n) for small values of n:
Look at case 1, with a = b, c = a + 1, and let k = abc - c = a^2(a+1) - (a+1) = (a-1)^2(a+1) and r = f^k(abc). Then we have the equation
2f(r) + r = a + b + c = 3a + 1. For instance, with a = 2, there has to be some natural number, r, that satisfies 2f(r) + r = 7, and the only possible candidates are 1, 3 and 5. Therefore:
Either f(1) = 3 or f(3) = 2 or f(5) = 1.
By putting a = 3, we see that either f(2) = 4 or f(4) = 3 or f(6) = 6. However, we already know that f(m) never equals m (Else we'd have 3abc = a +b + c for m = abc), so we exclude the last of these possibilities.
Next, case 2, with b = c = a+1. Again, with r = f^k(abc) and, this time, k = abc - a = a(a+1)^2 - a we get the relation: 2r + f(r) = a + b + c = 3a + 2. Whence, setting a=2, we see that:
Either f(1) = 6 or f(2) = 4 or f(3) = 2
Moving on to case 3, with b = a-1, c = a+1 and k = abc-c = a(a^2-1)-a-1, we get r + f(r) + f^2(r) = a + b + c = 3a. The "lowest" case has a=3, so that either f(1) + f^2(1) = 8 or f(2) + f^2(2) = 7 or f(3) + f^2(3) = 6 or f(4) + f^2(4) = 5 or f(5) + f^2(5) = 4 or f(6) + f^2(6) = 3
Edited on August 21, 2021, 5:53 pm
|
|
Posted by FrankM
on 2021-08-21 13:17:18 |
Further thoughts (4)
