Background:
Solving this could lead to something revolutionary which is why I don't think it is possible ... but it seems feasible. Despite the seemingly simplistic question, it is actually quite difficult and gets involved in a lot of number and series theory which I hint at as you keep reading.... I've been working on this for years (very part-time albeit) and I've finally given up.
Well-Formatted Question:
Is it possible to calculate with complexity O(1) the first point where two series align wherein one series is the square of all integers and the other is governed by the same growth (specifically the growth of $a_n$ where $a_n$ = $a_{n-1}$ + 2) with a starting point C? Basically, calculating the next point in the intersection of two arithmetic progressions, but if the two series only have one intersection, that is fine too.
Example:
Bunny Charles is at starting position 13. He first jumps 21 spaces forward, then 23, 25 etc... He is trying to land a perfect square (a number that can be square-rooted and result in an exact integer)... without just guessing, show how at which square he will first land on 'or' how many hops it takes to get there.
De-composing the question:
Charles starts at C=13, jumps with $a_n$ where $a_n$ = $a_{n-1}$ + 2 and $a_n$ is initially 21. ... knowing he is trying to land on the square of some number reveals that it is comparing two series where the other series is just squares: 1, 4, 9, 16, etc. It is well-known how the difference between squares holds to the same growth function (e.g. (diff 1,4 = 3, diff 4,9 = 5, diff 9,16 = 7 etc...).
Answer - Cheating:
Charles's hops are like this: 13, 34, 57, 82, 109, 138, 169 - ah ha! 169 is 13 squared.
Answer - Not Cheating:
Is there a way to predict that it will be 169 or 13 squared without just running the series like I did? Or, is there a way to predict the number of hops to get to 169 (ans: 6). I'm not going to show it, but answering either one could lead to the answer of the other with a time complexity of O(1) - constant time, meaning you do some calculation on the numbers and won't have to do additional steps for larger series (or where there happen to be more hops before a square is landed). Unless it only takes one hop and the answer happens to just be stupidly simple, the only complexity change for different series is that series with larger numbers means computations will be done on larger numbers, but that can still be O(1).
Interesting Thinking Points:
Series of hops: 13, 34, 57, 82, 109, 138, 169
Differences of hops: 21, 23, 25, 27, 29, 31
Series of Related Squares: 100, 121, 144, 169, 196, 225, 256
Note the relationship between the two series above.
Once that is understood, it is interesting that if you continue the hops until the difference in the two series is the same as the difference between two consecutive squares (e.g. $43^2$ and $44^2$), you are getting another point where Charles lands on a square. In other words, Charles lands on a square which happens to be 1 square smaller than the square represented by the growth in his last hop. What is that value and why is this?
The value is 1849, just after Charles hops 87 spaces from 1762. 1849 is 43x43. The growth/hop of 87 is the difference between 1849 and the next square -- 1936, or in other words, $43^2$ and $44^2$. After this point, Charles will be hopping over perfect square points forever. Or, in other words, there will never be any more points where the bunny lands on a square. This is because of a property of the growth rates and the rabbit will be stuck between two squares for every future hop beyond his hop of 87 spaces (which is his 34th hop <- (44 - 10 where 10 comes from the growth between 10 squared and 11 squared which is 21 ... which is Charles's initial growth ... you could also view this as 43 - 9)). However, don't let this section distract you, we need to know where Charles lands his first square, not his second...
Closing Thoughts:
Due to some interesting properties, with a different starting position and/or growth rate, you may find that the rabbit will only land on one perfect square (which would be governed by the reasoning of "Interesting Thinking Points" above - Charles's second hop in this case.) Or, you may find that multiple squares in the series can be aligned (3, 4 ...).
I've been given a link to the Chinese remainder theorem here - https://en.wikipedia.org/wiki/Chinese_remainder_theorem - however, I'm not sure how this can be used as a computational way to solve this problem, only methods to solve the problem, but I might not be understanding it right, so please provide input if you know of these theorems. Or, perhaps re-arranging with some algebra can arrive at a solution...
If this is impossible, provide a proof that it is impossible. If there is a solution, please give the solution.
If this has been answered already, even partially, feel free to provide links to those resources.
After attempting the problem, see the comments below for why this would be "revolutionary" - note - I have extracted this problem from the revolutionary problem just to focus the problem given here (since you don't need to know the revolutionary problem to solve this one - rather, the other way around perhaps).
I will assume based on your examples that the first hop is odd, so all hops are odd. If the first hop is of size $2k+1$ we can imagine that there have already been $k-1$ hops at the start, of sizes $1,3,5,\ldots 2k-1$ with total $k^2$. If the starting point is $b$ after $m$ hops we are at $b-k^2+(k+m)^2$. You are asking to find the minimum $m$ given $b,k$ that makes this a perfect square. If this is $n^2$ we have $$b-k^2+(k+m)^2=n^2\\ b-k^2=n^2-(k+m)^2=(n-k-m)(n+k+m)$$ You need to factor $b-k^2$ into two factors of the same parity. If $b-k^2 \equiv 2 \pmod 4$ there is no solution. Factoring is not $O(1)$ so this does not satisfy your request, but it is easier than brute force search. You want the two factors to be as close to $2k$ apart as possible among the factorizations of $b-k^2$.