multiplying by $4$ gives that $4(n^2+r-k(k+1))=4n^2+4r+1-(2k+1)^2$ does not equals to $-4$ or $4 \times$ (composite number) let $4n^2+4r+1=T$, obviously $T \equiv 1(mod 4)$ If $T=ab$ for some odd integer $b>a\ge3$, then $k=\frac{a-1}{2}$ gives $T-(2k+1)^2=a(b-a)$. $a \frac{b-a}{4}$ is not $-1$ or composite number Since $\frac{b-a}{4}$ is integer ($T \equiv 1(mod 4)$ gives $b-a \equiv 0 (mod 4)$.), $a=1$ or $\frac{b-a}{4}=1$. But $a>3$, so we get $b-a=4$. $T=ab=a(a+4)=a^2+4a$, then $k=\frac{a+1}{2}$ gives $T-(2k+1)^2 =-4$, contradiction. If $T=p^2$ for some odd prime $p$, $k=1$ gives $T-(2k+1)^2=p^2-9=(p-3)(p+3)$. $\frac{(p+3)(p-3)}{4}$ is not composite, so $p=3$. This gives $T=9$ Thus, $T=9$ or $T$ has at most one prime divisor. So, possible values of $T$ is $1, 9, p$ where $p$ is prime