1. ## Geometric 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?

2. I am sorry if i just dont understand, but could you please be a little more clear?

3. 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?
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?

4. *distinct* integers, so a not equal to b

5. 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.)

6. 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

7. 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.

8. 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?

9. Well then im working in reverse but who cares...

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.

10. 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.

11. 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

12. I can't follow the reasoning used in this bit:

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?

13. Here.

(See where you can spot where I use the fact that they are distinct in the proof).

14. 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?
Ah. I missed that case. Oh well, it WAS 3:30 in the morning after all...

-Dan

15. 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.)