# Thread: A number theoritic problem

1. ## A number theoritic problem

Please help me with a solution to the following problem,I've been trying this for somedays but is unable to approach it in any way...
N1=2, N2=3, N3=5, N4=6 , ... , Ni=i-th non square integer.

It is found that for some integer m, m^2< Nn <(m+1)^2

Prove that m= [√n + (1/2)] where [x]=Greatest Integer Function.

2. Originally Posted by Sarasij
N1=2, N2=3, N3=5, N4=6 , ... , Ni=i-th non square integer.

It is found that for some integer m, m^2< Nn <(m+1)^2

Prove that m= [√n + (1/2)] where [x]=Greatest Integer Function.

The number of squares between 2 and $\displaystyle N_n$ is $\displaystyle k_n:=[\sqrt{n}]-1$ (why?) , so $\displaystyle N_n=n+k_n$ ,

and thus $\displaystyle m^2<N_n<(m+1)^2\Longrightarrow$...(complete)

Tonio

3. Originally Posted by tonio
The number of squares between 2 and $\displaystyle N_n$ is $\displaystyle k_n:=[\sqrt{n}]-1$ (why?) , so $\displaystyle N_n=n+k_n$ ,

and thus $\displaystyle m^2<N_n<(m+1)^2\Longrightarrow$...(complete)

Tonio
Perhaps it is $\displaystyle N_n=n+k_n+1$ instead $\displaystyle N_n=n+k_n$ ?

If not, can you show some calculations please?

4. Originally Posted by Also sprach Zarathustra
Perhaps it is $\displaystyle N_n=n+k_n+1$ instead $\displaystyle N_n=n+k_n$ ?

If not, can you show some calculations please?

Of course, it is $\displaystyle N_n=n+k_n+1$ . Thanx

Tonio

5. ## Re: A number theoritic problem

Thanks a lot guys...I've solved it in another way 2...
Nn=n+1 , 1<= n <=2
=n+2 , 3<= n <=6
=n+3 , 7<= n <=12...
=n+k , k^2 - k + 1 <= n <= k^2 + k

From here a little simplification tells that "(√n + 1/2)-1 < m < √n + (1/2)"
which clearly implies the proof...

6. ## Re: A number theoritic problem

Originally Posted by Sarasij
Thanks a lot guys...I've solved it in another way 2...

No, you didn't. The following is exacty the same formula shown to you above.

Tonio

Nn=n+1 , 1<= n <=2
=n+2 , 3<= n <=6
=n+3 , 7<= n <=12...
=n+k , k^2 - k + 1 <= n <= k^2 + k

From here a little simplification tells that "(√n + 1/2)-1 < m < √n + (1/2)"
which clearly implies the proof...
.

7. ## Re: A number theoritic problem

I dont get it. u are dealing with no of squares. but i investigated the range of n for the condition k2<n<(k+1)2 to hold true. plz explain.

8. ## Re: A number theoritic problem

okk... the k factor in my calculations involves the k_n you have shown. yours is more generalised. can u show the proof of the statement that k_n=[√n]-1?

9. ## Re: A number theoritic problem

Originally Posted by Sarasij
okk... the k factor in my calculations involves the k_n you have shown. yours is more generalised. can u show the proof of the statement that k_n=[√n]-1?

Well, I think that is pretty straightforward and I leave the easy proof to you. Please note that the -1 is there because we don't count 1...

Tonio