• May 4th 2007, 12:06 PM
icebag
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 PM
janvdl
I am sorry if i just dont understand, but could you please be a little more clear?
• May 4th 2007, 12:16 PM
ThePerfectHacker
Quote:

Originally Posted by icebag
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:

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 PM
icebag
*distinct* integers, so a not equal to b
• May 4th 2007, 12:40 PM
icebag
Quote:

Originally Posted by janvdl
I am sorry if i just dont understand, but could you please be a little more clear?

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 PM
janvdl
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 PM
ThePerfectHacker
Quote:

Originally Posted by janvdl
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

You cannot do that.
Because you are assuming that GM and QM are equal.

They are not necessarily equal.
• May 4th 2007, 01:07 PM
janvdl
:) 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 PM
janvdl
Well then im working in reverse but who cares...:D

Quote:

Originally Posted by janvdl
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

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 PM
janvdl
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 PM
topsquark
Quote:

Originally Posted by icebag
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.)

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 AM
icebag
I can't follow the reasoning used in this bit:

Quote:

Originally Posted by topsquark
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.

The product (2x + 1)(2y + 1) can be a perfect square without either term being a factor of the other: e.g. 9 x 25 has square root 15. Am I missing something?
• May 5th 2007, 05:12 PM
ThePerfectHacker
Here.

(See where you can spot where I use the fact that they are distinct in the proof).
• May 5th 2007, 06:16 PM
topsquark
Quote:

Originally Posted by icebag
I can't follow the reasoning used in this bit:

The product (2x + 1)(2y + 1) can be a perfect square without either term being a factor of the other: e.g. 9 x 25 has square root 15. Am I missing something?

:o Ah. I missed that case. Oh well, it WAS 3:30 in the morning after all...

-Dan
• May 6th 2007, 02:03 AM
icebag
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.)