# Conjecture Time!

1. Jun 30, 2013

### Tenshou

I don't know if this is the proper thing to call it, but I haven't used any mathematical terminology in a while so I think I will try :P

The number of imperfect roots between any two consecutive perfect roots will always be twice the preceding root number.

for example there is 2 imperfect roots between $\sqrt1 and \sqrt4$ and 6 in between $\sqrt9 and \sqrt16$ This pattern seems to work I have tried this up for large numbers and it seems to hold, it seems like it should very easy to prove and I am working on this proof as we speak, but I think I could use a little help...

2. Jun 30, 2013

### Staff: Mentor

Hint: if n^2 is a perfect square, the next perfect square is (n+1)^2. How many integers (edit: better) are between those two numbers?

Last edited: Jun 30, 2013
3. Jun 30, 2013

### Tenshou

I don't understand the question mfb? there are countably infinite number, correct?

My question is strictly for natural number D:

4. Jun 30, 2013

### Staff: Mentor

I mean integers only, of course.

5. Jun 30, 2013

### Tenshou

Well, then it would always be a odd number. unless you count the number you start on then it should be even every time?

6. Jun 30, 2013

### Staff: Mentor

How is 2 (as an example, the number of integers between 1 and 4) odd?
You can expand (n+1)^2 and get the number of integers between n^2 and (n+1)^2...

7. Jun 30, 2013

### Tenshou

Sorry, I was counting like this 1...2*...3*...4*, I counted each element, like the distance >.< LOL! But you are right that is true, so it works for both squares and square roots, oh that is interesting, I should have checked that case... but now the real question is "Can I make a generalization for this for the n-th root of any "consecutive" numbers like$\sqrt[n]m$ and $\sqrt[n]k$ are there n times the previous number where m and k are integers?" But I think you answered my question.

8. Jun 30, 2013

### Mandelbroth

Let $n$ be a positive integer. You are looking to see if $2n+1 \stackrel{?}{=} (n+1)^2-n^2$.

$$2n+1 \stackrel{?}{=} (n+1)^2-n^2 \\ 2n+1 \stackrel{?}{=} ((n+1)-n)((n+1)+n) \\ 2n+1 = 2n+1$$

Last edited: Jun 30, 2013
9. Jun 30, 2013

### Staff: Mentor

Not n times the previous number, but you can get similar (just more complicated) formulas with the same approach as for the square root (n=2).

10. Jun 30, 2013

### Tenshou

So, they become more complicated than just n times the previous number? I mean it works for consecutive square roots of perfect squares, and it would be a little more complicated for consecutive cubes of perfect cubes and all the way forward, This is awesome!

11. Jul 1, 2013

### Staff: Mentor

Just calculate some examples, and you'll see it.

12. Jul 1, 2013

### Tenshou

I did, and thanks, so I guess it busted that hypothesis lol