Perfect‑Square Discriminant / Factor‑Pair Technique

Use the discriminant condition for integer roots.
For $x^{2}-kx+16=0$,
$$ D = k^{2}-4\cdot16 = k^{2}-64 $$
must be a positive perfect square.
So there exists an integer m>0m>0 such that
$$ k^{2}-64 = m^{2}\quad\Longrightarrow\quad k^{2}-m^{2}=64. $$
Factor as a difference of squares.
$$ (k-m)(k+m)=64. $$
Translate to factor pairs of 64.
Let
$$ s = k-m,\quad t = k+m, $$
with $s,t = 64$ and both $s,t>0$.
Because $k=\tfrac{s+t}{2}$ and $m=\tfrac{t-s}{2}$ must be integers, $s$ and $t$ must be the same parity (both even, since 64 is even).
List the positive even factor pairs $(s,t)$ of $64$:
$$ (2,32),;(4,16),;(8,8). $$
(Pairs where $s>t$ would simply swap $m$’s sign; we only need each unordered pair once.)
Compute $k$ for each pair.
$$ \begin{aligned} (s,t)=(2,32)&:&;k=\tfrac{2+32}{2}=17,\[4pt] (s,t)=(4,16)&:&;k=\tfrac{4+16}{2}=10,\[4pt] (s,t)=(8,8)&:&;k=\tfrac{8+8}{2}=8\quad\text{(but then }m=0,\text{ gives a double root).} \end{aligned} $$
We need distinct roots, so discard $k=8$.
Answer.
The admissible values are $k=10$ and $k=17$.
Hence there are
$$ \boxed{2} $$
positive integers $k$ producing distinct integer roots.

Perfect‑Square Discriminant / Factor‑Pair Technique#