For odd positive integer m=c^{2}-r, define the characteristic polynomial f(x)=x^{2}+2cx+r. We are looking for the least nonnegative integer x for which f(x) is a perfect square.

Completing the square of the polynomial may give us a way to find values of x given f(x). I decided to try to determine when f(x) = 0, 1, 4, 9, and so on, so that I would have a series I could examine to try to predict when x would be an integer.

f(x)=(c+x)^{2}-m, it turns out. When f(x)=a^{2} for some a (not necessarily integer, x=(m+a^{2})^{1/2}-c. With c an integer, we can discount it, and look at the series of the square roots of m+a^{2} to watch how the numbers behave. x will not be an integer until m+a^{2} is a perfect square.

For a small scale example, take m=689. c=27. 689+0^{2}, the portion of the expression of x under the radical for f(x)=0, lies between 26^{2} and 27^{2}. x stays between the squares of 26 and 27 for a while as a increases.

`0: 689+0 = 689 = 26`^{2} + 13 = 27^{2} - 40.
1: 689+1 = 690 = 26^{2} + 14 = 27^{2} - 39.
2: 689+4 = 693 = 26^{2} + 17 = 27^{2} - 36.
3: 689+9 = 698 = 26^{2} + 22 = 27^{2} - 31.
4: 689+16 = 705 = 26^{2} + 29 = 27^{2} - 24.
5: 689+25 = 714 = 26^{2} + 38 = 27^{2} - 15.
6: 689+36 = 725 = 26^{2} + 49 = 27^{2} - 4.

This is the last value of f(x)=a^{2} in the series for which x will be between 26 and 27. If, for a=7, we had hit the boundary value 27^{2} exactly, the progression would have been quadratic and uninterrupted. But now there is the change of “betweenness” to deal with.

This jump in the perfect squares between which a value lies is typical of the jumps encountered in any of the various methods I have explored to find a formula to predict when a value I am trying to track hits the perfect square boundary exactly. I plan to explore it more, and hope I will see what I have not yet seen, in terms of a way to anticipate that value, not only algorithmically, but with a closed formula.

It’s not over yet!

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