# Finding the Upper and Lower Bounds of Possible $x$ values

Given that $$n$$ is an odd semiprime (not a square of primes), $$a=\operatorname{ceil}(\sqrt{n})$$ and $$b=a^2-n$$. We also have a function, $$f(x)$$, such that $$f(x)=\frac{b-x^2}{2x-2a}$$ For each $$n$$, there is one trivial $$x$$ value that is equal to $$x=a-1$$ For each $$n$$, there is also one non-trivial positive $$x$$ value that is an integer (not including $$0$$) that also makes $$f(k)$$ a positive integer (including $$0$$). We'll call this $$k$$. What I am asking is this: what are the tightest bounds that this integer $$k$$ must be within? The ones I have is: $$-1+\sqrt{1+b+2a}\leq k\leq a-2$$ I don't know if those are the tightest. Also, what I want to know is what are the bounds that $$f(k)$$ must be within? All bounds must be in terms of $$a,b$$ and $$n$$.

The initial premise is wrong. In fact, there are precisely eight integers $$k$$ such that $$f(k)$$ is an integer (including $$a-1$$).

For this $$k$$ we have $$2(k-a)\mid b-k^2$$. But we have $$b-k^2\equiv b-a^2=-n\pmod{k-a}$$, therefore $$k-a\mid n$$. If we write $$n=pq$$ with $$p,q$$ odd primes, then we have $$k-a=\pm 1,\pm p,\pm q,\pm pq$$. And, indeed, any of those values can appear.

Let $$k$$ be any of the numbers $$a\pm 1,a\pm p,a\pm q,a\pm pq$$. For each of those $$k-a$$ is odd and, by going with the previous calculation backwards, $$k-a\mid b-k^2$$. Further, $$k$$ has parity opposite to that of $$a$$ (if $$a$$ is even, $$k$$ is odd and vice versa), and the same is true of $$b$$, so $$b-k^2$$ is even. It follows that $$2(k-a)\mid b-k^2$$, and hence $$f(k)=\frac{b-k^2}{2k-2a}$$ is an integer.

Edit: Let me assume $$p. Then of the eight values of $$k$$ above all are positive, except for $$a-q,a-pq$$. To get $$f(k)$$ positive, we must either have $$k>a$$ and $$b\geq k^2$$ or $$k and $$b\leq k^2$$. The former is impossible, since it implies $$b\geq (a+1)^2$$, and this is easily seen to never hold - indeed, $$b\leq 2a-1$$.

Therefore we have $$k. This implies that $$k=a-1$$ or $$k=a-p$$, so $$a-p$$ is necessarily the value you seek for. In terms of $$n$$, you can't say anything more about $$p$$ than $$3\leq p\leq\sqrt{n+1}-1$$ (which happens when $$q=p+2$$), so $$a-\sqrt{n+1}+1\leq k=a-p\leq a-3$$. If you want a bound in terms of just $$a,b$$ you get $$a-\sqrt{a^2-b+1}+1\leq k\leq a-3$$ and those bounds are optimal.

• Whoops meant to say that $f(k)$ is positive. – Quote Dave Jun 24 at 19:27
• $k$ must also be positive. – Quote Dave Jun 24 at 19:28
• @QuoteDave In that case, $k$ may not exist, take e.g. $n=15$. – Wojowu Jun 24 at 19:40
• Since $b=0$, $f(1)=0$ which is not positive. – Wojowu Jun 24 at 19:43
• $0$ is usually (in the English-speaking countries at least) not considered to be positive nor negative. Rather, it is nonnegative and nonpositive. – Wojowu Jun 24 at 19:45