I suspect that there is no pair of distinct integers a, b for which the geometric mean and the quadratic mean (root mean square) are both integers, but I am getting nowhere trying to prove it. Any ideas?
By Quadradic Mean I think you mean, Generalized Power Mean which exponent 2.
Then if you choose a=b integers.
Then,
QM(a,b)=a=b --> integer.
GM(a,b)=a=b --> integer.
What is the problem?
This is as far as i get...
g = sqr(ab)
q = sqr(0,5a^2 + 0,5b^2)
g = q
sqr(ab) = sqr(0,5a^2 + 0,5b^2)
ab = 0,5a^2 + 0,5b^2
2ab = a^2 + b^2
0 = a^2 - 2ab + b^2
0 = (a - b)^2
Well then im working in reverse but who cares...
So if we take the sqr root on both sides:
0 = a - b
a = b
So for g to equal q:
a must equal b.
So if a and b are distinct, my logic tells me that g will not equal q.
I dont know if everything is right, but maybe it can give you a clue.
I can reduce these conditions to a single equation. Perhaps ThePerfectHacker can take it from where I have to leave it.
Let b > a. Then b = a + n where n is some positive integer. Then
q = sqrt{(1/2)*(a^2 + a^2 + 2an + n^2)}
q = sqrt{a^2 + an + (1/2)n^2}
For this to be an integer n must at least be an even number. Thus a and b are either both even, or both odd.
Case 1: a, b are both odd.
Let a = 2x + 1
Let b = 2y + 1
(x, y both integers)
Then
q = sqrt{(1/2)(4x^2 + 4x + 1 + 4y^2 + 4y + 1)}
q = sqrt{2(2x^2 + 2x + 2y^2 + 2y + 1)}
But for q to be an integer 2x^2 + 2x + 2y^2 + 2y + 1 = 2(x^2 + x + y^2 + y) + 1 must be an even number, which is false.
Thus a, b are not both odd.
Case 2: a = 0 (mod 4) and b = 2 (mod 4)
Let a = 4x
Let b = 4y + 2
(x, y both integers)
Then
q = sqrt{(1/2)[(4x)^2 + (4y + 2)^2]}
q = sqrt{2(4x^2 + 4y^2 + 4y + 1)}
But for q to be an integer 4x^2 + 4y^2 + 4y + 1 = 2(2x^2 + 2y^2 + 2y) + 1 must be an even number, which is false.
Thus etc.
Case 3: a = 0 (mod 4) and b = 0 (mod 4)
Let a = 4x
Let b = 4y
Then g = 4*sqrt{xy}
Then q = 4*sqrt{(1/2)(x^2 + y^2)}
If g and q are to be integers then this implies that the geometric mean and quadratic means of x and y are integers. This is to say that we may divide a and b by 4 and still have a pair of numbers that works. So this is not a case in and of itself, but can always be reduced to one of the other cases. Since no other case works, a and b must then be of the form...
Case 4: a = 2 (mod 4) and b = 2 (mod 4)
Let a = 4x + 2
Let b = 4y + 2
(x, y are integers)
Then
g = sqrt{(4x + 2)(4y + 2)} = 2*sqrt{(2x + 1)(2y + 1)}
The condition that g is an integer implies that 2x + 1 is a factor of 2y + 1 of the form:
2y + 1 = (2x + 1)*z^2
where z is some integer.
Thus
q = sqrt{(1/2)[(2x + 1)^2 + (1/2)[(2x + 1)z^2]^2}
q = (2x + 1)*sqrt{(1/2)(z^4 + 1)}
If q is to be an integer then so is c = sqrt{(1/2)(z^4 + 1)}.
Thus we need to solve the equation:
2c^2 - z^4 = 1 where c, z are integers.
If a solution to this equation exists we may pick a value of x and then generate a value for y, and thus a and b.
I can't solve this (or prove it has no solution). Perhaps TPH can do it since he seems to have studied these types of equations.
-Dan