# For which natural numbers n does the expression

1. Aug 23, 2008

### dodo

A nice puzzle I just found (hope it hasn't been posted before):

For which natural numbers n does the expression
$$\sqrt {30 + \sqrt n} \ \ + \ \ \sqrt {30 - \sqrt n}$$​
yield also a natural number?

2. Aug 23, 2008

### CRGreathouse

Re: Puzzle

Well, it's a bounded expression... just check 0 to 900. ;)

3. Aug 23, 2008

### dodo

Re: Puzzle

Lol - I meant, with some fun explanation as of why.

4. Aug 23, 2008

### snipez90

Re: Puzzle

Well I know I've seen problems that ask you how many values of x are there such that

$$\sqrt{a - \sqrt{x}}$$ is an integer, where a is a constant.

This is essentially the same problem except you don't really need to worry about the first term. Just keep in mind the inequalities that must be satisfied and it just comes down to finding squares.

5. Aug 23, 2008

### dodo

Re: Puzzle

Well, the interesting thing is that the values of x for which 30-sqrt(x) is a square, will not produce also squares for 30+sqrt(x).

So this is a case where the two big roots are not integers, yet their sum is.

6. Aug 23, 2008

### snipez90

Re: Puzzle

Ahhh you're right. I totally reduced the problem to a simpler one without thinking. Thanks for pointing that out. I'll try to find a systematic solution.

7. Aug 24, 2008

### uart

Re: Puzzle

Let $$z = \sqrt {30 + \sqrt n} \ \ + \ \ \sqrt {30 - \sqrt n}$$

then $$z^2 = 60 + 2\, \sqrt{900 - n}$$

hence z^2 - 60 is an even natural number less than or equal to 60.

Listing the possible values of z^2 - 60 gives :

8^2 - 60 = 4
10^2 - 60 = 40
and they are the only possibilities.

So $2\, \sqrt{900 - n}$ equals either 4 or 40 and the corresponding values of n = 896 or n = 500 are easily calculated.

Last edited: Aug 24, 2008