Discover Patterns in Finite Series: Solving for n with 2008 Square Roots"

[arpitm08]

Homework Statement

Are there positive integers m and n so that
n = sqrt(m+sqrt(m+sqrt(m+...sqrt(m))))
where there are 2008 square root signs.


2. The attempt at a solution

n^2=m+sqrt(m+sqrt(m+sqrt(m+...sqrt(m))))...with 2007 square roots.
Idk where to go from here. If there were infinite roots, then it would be a lot easier to do.
Any suggestions would be helpful. Thanks.

 

[flatmaster]

My first instinct is to get rid of ugly square roots. Square both sides and subtract m 2008 times to get a 2008th degree polynomial in m. You may be able to somehow write this as a binomial.

 

[arpitm08]

I don't see where that would get me tho.

((((n^2-m)^2-m)^2-m)^2...-m)^2-m=0

Where do i go from there. Its still as complicated as before.

 

[Dick]

How about this? If n=sqrt(m+a) with n and m integers, then a is an integer. That tells you that the expression with 2007 square roots is an integer. Etc. Etc. Eventually you get down to sqrt(m+sqrt(m)) is an integer. Is that possible?

 

[arpitm08]

I don't think its possible, b/c ur assuming that m and n are integers.

 

[Dick]

I don't think its possible, b/c ur assuming that m and n are integers.

Can you show it's not possible? You are talking about k=sqrt(m+sqrt(m)), I assume.

 

[arpitm08]

So it would become:

((((a^2-m)^2-m)^2-m)^2...-m)^2=m

Are u saying that since a and m are integers, this expression would be an integer too, which would equal m??

 

[Dick]

No. I'm saying if n=sqrt(m+a) that means 'a' must be an integer. Where 'a' is the expression with 2007 square roots. Then the integer a=sqrt(m+b), where b is the expression with 2006 square roots. b must also be an integer so b=sqrt(m+c) where c is the expression with 2005 square roots. Etc etc. Eventually you get down to z=sqrt(m+sqrt(m)) is an integer. Is THAT possible?

 

[arpitm08]

YES! And then would you say that since sqrt(m+a) is an integer, there exists integers that satisfy this equation??

 

[Dick]

No, again. Because I don't think you can solve that final equation, z=sqrt(m+sqrt(m)) where z and m are integers. Try it.

 

[arpitm08]

Thank You Very Much! =)