7. Let $p$ be a given odd prime, if the positive integer $k$ makes $\sqrt{k^{2}-p k}$ also a positive integer, then $k=$
7. $\frac{(p+1)^{2}}{4}$ Detailed Explanation: Let $k^{2}-p k=a^{2}, k^{2}-p k-a^{2}=0$,

then $\Delta=p^{2}+4 a^{2}=m^{2}$, i.e., $p^{2}=(m-2 a)(m+2 a), \therefore m-2 a=1, m+2 a=p^{2}$. $a=\frac{p^{2}-1}{4}, m=\frac{p^{2}+1}{2}, k=\frac{(p+1)^{2}}{4}$.