Has this question ever been posed in the literature?
cafe...@msn.com
Let x_1,x_2,...,x_n be a sequence of positive reals. Both the
arithmetic mean and the geometric mean of this sequence are positive
integers.
Does this imply that each x_1,x_2,...,x_n is a positive integer?
Anyone know the answer?
Thank you,
Craig
WM
Here is a counter example:
6 - sqrt(11), 6 + sqrt(11)
arithmetic mean = 6
geometric mean = 5
There are many.
Regards, WM
Lee Rudolph
cafe...@msn.com writes:
When you write "Let x_1,x_2,...,x_n be a sequence of positive
reals. Both the arithmetic mean and the geometric mean of this
sequence are positive integers", do you mean that
"x_1,x_2,...,x_n" is to be an infinite sequence of positive
integers (with general term x_n), and that the two means are
to be positive integers for every n?
If so, I have no idea (about the answer; I *do* have the
idea that you should--in that case--have written more
precisely).
If not--that is, if your hypotheses could have been prefaced
with "Let n>1 be a positive integer."--then the answer is
"no" already for n=2; the first of an obvious infinite
family of counterexamples is x_1=4-sqrt(7), x_2=4+sqrt(y).
By self-concatenation, any counterexample x_1,...,x_n of
length n gives a counterexample of length 2n.
Lee Rudolph
WM
> arithmetic mean and the geometric mean of this sequence are positive
> integers.
>
> Does this imply that each x_1,x_2,...,x_n is a positive integer?
>
For sequences of four terms we obtain the condition
(a-b)(a-b)(a+b)(a+b) = (a^2 + b^2)^2 - 4a^2b^2 = k^4
with k a natural number and a-b a positive non-natural number.
Solutions are, for instance,
a = 2, b = sqrt(3), k = 1
a = 3, b = sqrt(5), k = 2
a = 4, b = sqrt(7), k = 3
and so on.
The number of terms can be increased too.
Regards, WM
Arturo Magidin
For n=2, x_1+x_2 = 2m and x_1x_2 = n^2 for integers n and m; so x_1
and x_2 are roots of the monic polynomial with integer coefficients
x^2 -(2m)x + n^2. The roots are integers if and only if the polynomial
is reducible over Q. The roots are m + sqrt(m^2-n^2) and m - sqrt(m^2-
n^2), so all you need is for m^2-n^2 to be a nonsquare. E.g., m=3,
n=2, to get x^2-6x+4, with roots 3+sqrt(5) and 3-sqrt(5); their
arithmetic mean is 3, their geometric mean is 2, neither is rational.
For n>2, pick your favorite nonzero x_1,...,x_{n-2}. You want to find
numbers a and b such that (x_1+...+x_{n-2}) + (a+b) is a multiple of
n, and (x_1*...*x_{n)*ab is an nth power. So you want a and b to be
roots of a polynomial of the form
y^2 - (nk-(x_1+...+x_n))y + m^n/(x_1*...*x_{n-2})
for some integers k and m. As long as the coefficients are not
integers, or if they are the resulting polynomial is not reducible
over Q, then the roots will not be integers. Seems like that would
occur for most choices of m and k, given nonzero x_1,...,x_{n-2}.
--
Arturo Magidin
serge bouc
Le 12.06.2010 15:12, cafe...@msn.com a �crit :
be a couterexample : for n=2, consider x_1=2+sqr(3) and x_2=2-sqr(3).
Agusti Roig
Take the sequence of the two positive real numbers
x_1 = 2 + \sqrt(3) , x_2 = 2 - \sqrt(3)
Agusti
Soquiso
tc...@lsa.umich.edu
<cafe...@msn.com> wrote:
>Let x_1,x_2,...,x_n be a sequence of positive reals. Both the
>arithmetic mean and the geometric mean of this sequence are positive
>integers.
>
>Does this imply that each x_1,x_2,...,x_n is a positive integer?
No. Let n = 2, x_1 = 2 - sqrt(3), x_2 = 2 + sqrt(3). The arithmetic
mean is 2 and the geometric mean is 1.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
William Elliot
Yes, I do for n = 2, it's no. Let
r = a + b
s = a - b
(r + s)/2 = a
sqr rs = sqr(a^2 - b^2)
Set b = sqr(2a - 1).
a^2 - b^2 = a^2 - 2a + 1 = (a - 1)^2
sqr rs = a - 1
Choose a in N to make sqr(2a - 1) irrational.
Since 0 < a - 1, for a > 1
2a - 1 < a^2; b < a; 0 < a - b < a + b
----
Heine Rasmussen
x_1 = 10+sqrt(51)
x_2 = 10-sqrt(51)
alainv...@gmail.com
Good morning Craig,
I propose a two variables example:
(x1+x2)/2 and (x1*x2)^(1/2)
We may choose x1 and x2 to obtain integer values:
Ex: x1=3-sqrt(5), x2 = 3+sqrt(5)
arithmetic mean = 3
geometric mean = 2
In general case Prod(xi) must be an exact n power number,
Alain
Martin Saturka
On Jun 12, 3:12?pm, cafei...@msn.com wrote:
> Let x_1,x_2,...,x_n be a sequence of positive reals. Both the
> arithmetic mean and the geometric mean of this sequence are positive
> integers.
>
> Does this imply that each x_1,x_2,...,x_n is a positive integer?
1/5, 1/5, 1/5, 1/5, 1/5, 5, 5, 5, 5, 5, 1, 1, 1, 1, 1, 1
M.
GJ Woeginger
cafe...@msn.com wrote:
#
# Let x_1,x_2,...,x_n be a sequence of positive reals. Both the
# arithmetic mean and the geometric mean of this sequence are positive
# integers.
#
# Does this imply that each x_1,x_2,...,x_n is a positive integer?
#
# Anyone know the answer?
An easy counterexample are the two roots x_1, x_2 of x^2-4x+9.
--Gerhard
___________________________________________________________
Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/
alainv...@gmail.com
On 13 juin, 15:35, gwo...@figipc78.tu-graz.ac.at (GJ Woeginger) wrote:
> cafei...@msn.com wrote:
>
> #
> # Let x_1,x_2,...,x_n be a sequence of positive reals. Both the
> # arithmetic mean and the geometric mean of this sequence are positive
> # integers.
> #
> # Does this imply that each x_1,x_2,...,x_n is a positive integer?
> #
> # Anyone know the answer?
>
>
> --Gerhard
>
> ___________________________________________________________
> Gerhard J. Woeginger ? ? ?http://www.win.tue.nl/~gwoegi/
Good morning,
x^2-4x+9 = (x-2)^2+5 doesn't work for real values.
But it gives us an idea:
why not considering n degree polynomials
x^n-a1*x^(n-1)+...+an ,
with all real roots and a1=np , an =q^n , p>=q (1)
a1, an ,p , q integer .
Example 1)
x^2 -6x +4 a1=6 =2*3 , a2 = 2^2
a1 = 3+sqrt(5) , x2 =3-sqrt(5)
Example 2)
x^4-16x^3+78x^2-144x+81
a1=4*4 , a4=3^4
roots {1,3,3,9}
What matters is to obtain only real roots polynomials
with conditions (1)
Any idea,
Alain
Chris Thompson
1/2, 1/2, 32 is simpler if you want to s/real/rational/
[There is no rational counter-example with n=2. Proof left as an
exercise for the reader.]
--
Chris Thompson
Email: ce...@cam.ac.uk
Gerry Myerson
In article
<hv2mi4$2mgt$1...@dhcp-128-146-115-56.osuwireless.ohio-state.edu>,
Lee Rudolph <lrud...@panix.com> wrote:
> cafe...@msn.com writes:
>
>
> >Let x_1,x_2,...,x_n be a sequence of positive reals. Both the
> >arithmetic mean and the geometric mean of this sequence are positive
> >integers.
> >
> >Does this imply that each x_1,x_2,...,x_n is a positive integer?
> >
> >Anyone know the answer?
>
> When you write "Let x_1,x_2,...,x_n be a sequence of positive
> reals. Both the arithmetic mean and the geometric mean of this
> sequence are positive integers", do you mean that
> "x_1,x_2,...,x_n" is to be an infinite sequence of positive
> integers (with general term x_n), and that the two means are
> to be positive integers for every n?
I don't think anyone has taken up this interpretation
(with the proviso that "infinite sequence of positive integers"
was supposed to be "infinite sequence of positive reals).
> If so, I have no idea (about the answer...
Well, here's a start:
3 + 2 sqrt 2, 3 - 2 sqrt 2, 27, 3, 729, ...
The next (= 6th) term, call it x, has to satisfy
7 divides x + 765 and 3^(10) x is a 7th power,
which is to say x = 5 mod 7 and x = 81 y^7
for some y.
I think it's clear that if there is such an infinite sequence
then all but its first two entries must be integers.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Arturo Magidin
On Jun 15, 7:15?am, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
wrote:
> In article
> <hv2mi4$2mg...@dhcp-128-146-115-56.osuwireless.ohio-state.edu>,
> ?Lee Rudolph <lrudo...@panix.com> wrote:
[...]
> > [D]o you mean that
> > "x_1,x_2,...,x_n" is to be an infinite sequence of positive
> > integers (with general term x_n), and that the two means are
> > to be positive integers for every n?
>
> I don't think anyone has taken up this interpretation
> (with the proviso that "infinite sequence of positive integers"
> was supposed to be "infinite sequence of positive reals).
>
> > If so, I have no idea (about the answer...
[...]
> I think it's clear that if there is such an infinite sequence
> then all but its first two entries must be integers.
This holds even if you only require the arithmetic mean to be an
integer. The geometric mean would play no role whatsoever.
Let r and s be the first two terms, so that r + s must be an even
integer. The third term t must satisfy r+s+t = 3k for some integer k,
hence t = 3k-2m for some integer m, thus t is an integer.
If you assume the first n terms are integers, so that x_1+...+x_n must
be equal to nk for some integer k, then x_{n+1} = (n+1)m - nk for some
integer m, hence must be an integer.
(Of course, if you truly require it for all n, and not merely all n>1,
then you also need x_1 to be an integer...)
--
Arturo Magidin
Martin Saturka
> Martin Saturka wrote:
> > 1/5, 1/5, 1/5, 1/5, 1/5, 5, 5, 5, 5, 5, 1, 1, 1, 1, 1, 1
>
> 1/2, 1/2, 32 is simpler if you want to s/real/rational/
Hello Chris,
the original question was on integer vs. non-integer reals. And non-
integer rationals are for sure non-integer reals. Generally, it is
nice to try to find as simple answers as possible. And rationals are
simpler than general reals.
The 1/2, 1/2, 32 is shorter than the 1/5 .. 5 .. 1 sequence, but the
latter one contains just p^i with i in -1,0,1.
If we would like to use sequences with elements a^i, i in -1,0,1, we
can have:
a-times a^-1, a-times a^1, and ((a-2)^2 - 3)-times a^0 for integer a
>= 4 (for both prime and non-prime a),
length is (a-1)^2, with the geometric mean = 1, and the arithmetic
mean = 2, proof is simple.
If we would like to use sequences with elements b^-1, b^j, j natural
number, we can have:
b-times b^-1, once b^(2*b+1), for any integer b >= 2 (again for both
prime and non-prime b),
length is b+1, with the geometric mean = 1, and the arithmetic mean =
\sum_{j=0}^{2*b} (-b)^j, i.e. it is a (1-2*q)-th q-integer for q = -b,
proof is via taking [b^(2*b+1) + 1] / (b+1) as [-((-b)^(2*b+1) - 1)] /
[-((-b) - 1)].
In fact, these arithmetic means form a nice integer sequence, with
taking the b=1 case too:
1, 11, 547, 52429, 8138021, 1865813431, 593445188743, ...
and it looks it is not contained in the integer sequence yet, see
http://www.research.att.com/~njas/sequences/?q=1%2C11%2C547%2C52429%2C8138021%2C1865813431%2C593445188743&sort=0&fmt=0&language=english&go=Search
M.