Challenge: converged limit of infinite sqrt(n)

Question

In math, sqrt(n+sqrt(n+sqrt(n+...))) has its limit. Interestingly, many limits of this kind are positive integers.

Write a code to find all "integer limit"s and their corresponding "n"s, within the input number "n". (i.e. from 0 to n)

The output should look like "n -- limit" or something like this.

I wonder that all the "n"s that match this "interger limit" condition may act as a arithmetic progression(A.P.).

sqrt=square root
example: 2 -- 2 ; 6 -- 3 ; 12 -- 4

Checking convergence may help.

Answer 1

The values of 'n' so that 'sqrt(n+sqrt(...))' is an integer are 0, 2, 6, 12, 20, ...

It's not an arithmetic progression. But observe the differences of consecutive terms
2, 4, 6, 8, ... which is an arithmetic progression.

Here's my try:
https://code.sololearn.com/Wr0q8lNC0tSL/?ref=app

Answer 2

https://code.sololearn.com/cs0DUKIThMAK/?ref=app

Answer 3

https://code.sololearn.com/c0DbJsjGnOMs/?ref=app

Answer 4

If your serie U(p+1)=sqrt(U(p)+n) has a limit l then l verifies l^2=l+n. By using the quadratic equation you can show that the positive solution is l=(1+sqrt(1+4n))/2. By expressing when this is equals an integer k you will solve your problem

Answer 5

no one gave the NEGETIVE LIMITS...

THERE WILL ALWAYS BE 2 LIMITS..
ONE (+ ) AND ONE( -)

https://code.sololearn.com/cEd8ivh94y04/?ref=app

https://code.sololearn.com/cV4CWKWeza52/?ref=app

this is another version simplified

"observe how the multiplicatipn of i and i-1 is the number
where i is the positive limit of that number"