Determine all possible triplets (x,y,z) of positive integers with x < y < z < 100 such that: x, y and z (in this order) are in arithmetic sequence and x, y and z+1 are in
harmonic sequence.
Bonus Question:
Amending the restriction to x < y < z, prove that there are an infinity of positive integer triplets satisfying the given conditions.
Let 1/x=1/(kb), 1/y=1/k, 1/z=1/(k+b). then 1/(kb)1/k = 1/k1/(k+b+1). This results in k=2(b^2+b) and we are in Triangular World.
e.g. 1/(41) 1/4 = 1/41/(4+1+1), {x,y,z} ={3,4,5}
1/(122) 1/12 = 1/121/(12+2+1), {x,y,z} ={10,12,14}
1/(243) 1/24 = 1/241/(24+3+1), {x,y,z} ={21,24,27} etc, continuing for {n,x,y,z}: {4,36,40,44}, {5,55,60,65}, {6,78, 84, 90}, when the next triplet puts z over 100.
for x, see A014105: n(2n+1) in Sloane and for z, A014106: n(2n+3) in Sloane.
Hence the substitution {x,y,z} = {2n^2+n, 2n^2+2n, 2n^2+3n} solves for all n, as is shown by:
1/(2n^2+n)1/(2n^2+2n) = 1/(2n(n+1)(2n+1)),
1/(2n^2+2n)1/(2n^2+3n+1) = 1/(2n(n+1)(2n+1)).
It also follows that 1/x and 1/(z+1) are the reciprocals of consecutive triangular numbers, the index of the first of which is even, and of the second odd.
x values are 1/(1*3), 1/(2*5), 1/(3*7), 1/(4*9) etc;
(z+1) values are 1/(2*3), 1/(3*5), 1/(4*7), 1/(5*9) etc.
z did not have to be 1/(k+b+1). Indeed, for z = 1/(k+b+x), x can be any positive integer:
For x>1, 1/(kb)1/k = 1/k1/(k+b+x): 2b^2+2bx = kx
So if x is not 1, then x divides b. Let b=c*x.
Now: 2(c*x)^2+2(c*x)*x = k*x
b=c*x, k=2c(c+1)x:
1/((2c(c+1)x)c*x)1/(2c(c+1)x) = 1/(2c(c+1)(2c+1)x)
1/(2c(c+1)x)1/((2c(c+1)x)+c*x+x) = 1/(2c(c+1)(2c+1)x)
Note that since the same 'triangular' formula appears: 2c(c+1) = 2c^2+2c, every solution will be a multiple of an (x=1) solution.
Edited on January 25, 2013, 10:52 am

Posted by broll
on 20130125 10:18:50 