Factors of a perfect square plus one

Question

For large integer $a$, small integer $d$, consider the following quantity:

$$a^2+d$$

What are the best lower bounds one can get for the sum $l+m$, where integers $l,m$ are such that:

$$lm=a^2+d$$

Trivially we have $2a+1$, can we do better?

Answer 1

Let the best lower bound sum of factors $l,m$ of number $lm=a^2+d$ be $l+m=2a-k$ for positive integers $a,d,l,m$ and integer $k$.

Since $l+m=2a-k,$ we have $l,m$ equidistant from ${2a-k\over2}=a-\frac k2$, so there exists an integer $c\equiv k\pmod 2$ such that $l=a-{k+c\over 2},m=a-{k-c\over 2}.$ Then we have

$$lm=\left(a-{k+c\over2}\right)\left(a-{k-c\over 2}\right)=\left(a-\frac k2\right)^2-{c^2\over 4}=a^2+d$$

$$a^2-ak+{k^2\over 4}-{c^2\over 4}=a^2+d$$ $$d={k^2-c^2\over 4}-ak\tag 1$$

There are two ways that $d$ can be positive due to $(1),$ which is to have $k\gt 4a$ or $k\lt 0.$ In the first case, we have $l+m \lt -2a$ which contradicts our assumption that $l,m$ are positive integers -- but also note that we then have $-l-m\gt 2a$. In the second case, we have $l+m\gt 2a,$ and the smallest integer value for this is $l+m=2a+1$ which is the claimed best lower bound by the trivial $l=a,m=a+1,d=a.$