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LordKinbote
Daedalian Member

 Posted: Tue Oct 02, 2012 11:47 pm    Post subject: 1 This is a variation I wrote on a problem I've seen in math competitions. What is the 100th positive integer a such that sqrt(a + sqrt(a + sqrt(a + sqrt(a + ...)))) is an integer? (I guess I could make an image of that, but I think you guys get the general idea: infinite nested square roots.)
groza528
No Place Like Home

 Posted: Wed Oct 03, 2012 4:43 am    Post subject: 2 With some quick math... 10100?
Chuck
Daedalian Member

 Posted: Wed Oct 03, 2012 4:59 am    Post subject: 3 That's what I got with a computer search.
Thok
Oh, foe, the cursed teeth!

 Posted: Wed Oct 03, 2012 6:16 am    Post subject: 4 In general, the nth number is n^2+n. The proof is not so hard (well unless you include the step where you prove the infinite square root makes sense and converges at all.
DejMar
(Possibly a robot)

 Posted: Wed Oct 03, 2012 10:36 am    Post subject: 5 The nth positive integer would be where a = (n^2 + n) with the expression sqrt(a + sqrt(a + sqrt(a + ...))) converging to the positive integer (n + 1). The 100th positive integer would be where a = (100^2 + 100) = 10100, with the expression converging to (100 + 1) = 101.
L'lanmal
Daedalian Member

Posted: Wed Oct 03, 2012 9:39 pm    Post subject: 6

 Thok wrote: In general, the nth number is n^2+n. The proof is not so hard (well unless you include the step where you prove the infinite square root makes sense and converges at all.

Um, those would be the things which make it a proof. Although this could be one of those "application to a mathematician is theory to an engineer" things. Try the word "solution". You might have a solution like: Find the points fixed by the iteration by noting that they occur at x = sqrt(a+x), note they are attractors (stable fixed points and not unstable ones) when viewing this as a 1-dimensional difference equation, Solve for a and note that a is an increasing function of x, and plug in 100 for x.

But hold on, when x=1, then a=0 isn't exactly a positive integer, is it? And everyone else is getting x^2+x rather than x^2-x. What's wrong?
lostdummy
Daedalian Member

Posted: Thu Oct 04, 2012 8:20 am    Post subject: 7

 L'lanmal wrote: But hold on, when x=1, then a=0 isn't exactly a positive integer, is it? And everyone else is getting x^2+x rather than x^2-x. What's wrong?

You stated reason in first part of that sentence ;p

Solution indeed is a=x(x-1), but when you want to index solution so that when you use n=1 you get first acceptable solution to problem (a>0), or when you use n=100 you get required solution, then you need to skip a=0, and thus you need to start at x=2.

So if we use substitution x=n+1, we get result where for n=1, a>0:
a=n(n+1)

If for example problem was 'find 100th number where a>10' then we would need to shift by 3, ie x=n+3, and thus a=(n+2)(n+3)
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