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? :confused:

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- May 4th 2007, 12:06 PMicebagGeometric and quadratic means
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? :confused:

- May 4th 2007, 12:12 PMjanvdl
I am sorry if i just dont understand, but could you please be a little more clear?

- May 4th 2007, 12:16 PMThePerfectHacker
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? - May 4th 2007, 12:19 PMicebag
*distinct* integers, so a not equal to b

- May 4th 2007, 12:40 PMicebag
Let a and b be unequal integers.

The geometric mean is g=sqrt(a*b).

The root mean square (or quadratic mean) is q=sqrt(0.5*(a^2+b^2)).

I want to prove that g and q cannot both be integers.

(Or, of course, find values of a and b such that g and q are both integers.) - May 4th 2007, 12:55 PMjanvdl
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 - May 4th 2007, 01:04 PMThePerfectHacker
- May 4th 2007, 01:07 PMjanvdl
:) Im sorry PerfectHacker, im not on varsity yet, im only trying to help.

But what if we do ASSUME them to be equal? Cant it aid us in finding a and b's value? - May 4th 2007, 10:23 PMjanvdl
Well then im working in reverse but who cares...:D

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. - May 4th 2007, 10:31 PMjanvdl
Hey icebag you edited your post...

The first one didnt say to prove that g and q were integers, but to prove them equal...

Ok, so my solution is way off, sorry. :D - May 4th 2007, 11:39 PMtopsquark
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 - May 5th 2007, 05:58 AMicebag
- May 5th 2007, 05:12 PMThePerfectHacker
Here.

(See where you can spot where I use the fact that they are distinct in the proof). - May 5th 2007, 06:16 PMtopsquark
- May 6th 2007, 02:03 AMicebag
Thank you TPH. That is very neat. :)

(Interestingly, I had got to the x^4 - y^4 = z^2 form myself, but I just didn't know that it had no integer solutions.)