For which natural numbers n does the expression

[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?

 

[CRGreathouse]

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

 

[dodo]

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

 

[snipez90]

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.

 

[dodo]

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.

 

[snipez90]

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.

 

[uart]

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?

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.