The von Mangoldt function $\Lambda$ is defined as follows: if $x=p^n$ is a prime power, then $\Lambda\left(x\right)=\mathrm{log}p$, otherwise $\Lambda\left(x\right)=0$. Let $P$ denote the set of all PRIME POWERS dividing $m$，let $S$ denote all the integer $x\in \left[c,c+\sqrt{m}\right]$ for which $m\mid x^2-a$ holds. WLOG assume that $\left(a,m\right)=1$ (If $d:=\mathop{\mathrm{gcd}}\limits_{s\in S}s$, then $\left(a,m\right)\mid d^2$ so we may substitute $\frac{a}{\left(a,d^2\right)},\frac{m}{\left(a,d^2\right)}$ for $a,m$.) Let $S\left(r,p\right)$ denote the number of elements in $S$ which $\equiv r\left(\mathrm{mod}p\right)$, then by Hensel Lifting, at most $2$ $S\left(r,p\right)$ are non-zero. Apply Cauchy-Schwarz, we get: $\left|S\right|^2=\left(\sum\limits_{r=0}^{p-1}S\left(r,p\right)\right)^2\leq 2\left(\sum\limits_{r=0}^{p-1}S\left(r,p\right)^2\right)=2\left(\left|S\right|+\sum\limits_{s_{i}\ne s_{j}\in S,s_{i}\equiv s_{j}\left(\mathrm{mod}p\right)}1\right)$ Summing over all $p\in P$, with weight $\Lambda \left(p\right)$: $\left|S\right|^2\sum\limits_{p\in P}\Lambda\left(p\right)\leq 2\left|S\right|\sum\limits_{p\in P}\Lambda\left(p\right) +\sum\limits_{s_{i}\ne s_{j}\in S}\sum\limits_{p\in P,p\mid s_{i}-s_{j}}\Lambda\left(p\right)$ Now we give an upper bound on the last summation on the right hand side above. Note that $\sum\limits_{d\mid n}\Lambda\left(d\right)=\mathrm{log}n$, so the summation $\leq \mathrm{log}\prod\limits_{s_{i}\ne s_{j}\in S}\left|s_{i}-s_{j}\right|$.By translation, we may assume $s_{i}\in\left[-\frac{\sqrt{m}}{2},\frac{\sqrt{m}}{2}\right]$(i.e., each $s_{i}$ is replaced by a $\tilde{s}_{i}$ with modulus no larger than $\frac{\sqrt{m}}{2}$); As $\prod\limits_{s_{i}\ne s_{j}\in S}\left|s_{i}-s_{j}\right|$ is exactly the square of the Vandermonde determinant, we may apply Hadamard's inequality (which claims $\left|\mathrm{det}A\right|\leq \prod\limits_{i=1}^{n}\left|\alpha_{i}\right|$, where $a_{i}$ denotes the row vectors of $A$), and deduce that $\prod\limits_{s_{i}\ne s_{j}\in S}\left|s_{i}-s_{j}\right|\leq \left|S\right|^{\left|S\right|}\left(\frac{\sqrt{m}}{2}\right)^{\left|S\right|\left(\left|S\right|-1\right)}$. So the inequality turns out to be $\left|S\right|\mathrm{log}2\leq\mathrm{log}m+\mathrm{log}k-\mathrm{log}\frac{\sqrt{m}}{2}\leq \mathrm{logm}+\mathrm{log2}$，which implies the conclusion.