let $n = m^2+k$ where as $ 0 \leq k \leq 2m $ thus $\lfloor{\sqrt{n}}\rfloor = m$ $m+\lfloor{\frac{m^2+k}{m}}\rfloor> 2\sqrt{m^2+k}$ $2m+\lfloor{\frac{k}{m}}\rfloor>2\sqrt{m^2+k}$ First Case : $\lfloor { \frac{k}{m}}\rfloor = 0$ $2m > 2\sqrt{m^2+k}$ $m^2>m^2+k$ implies $k <0$ however $k \geq 0$ so no such n Second Case: if $\lfloor {\frac{k}{m}\rfloor} = 1$ $2m+1 > 2\sqrt{m^2+k}$ $4m^2+4m+1 >4(m^2+k)$ $4m^2+4m+1>4m^2+4k$ $4m+1>4k$ $k< \frac{4m+1}{4}=m+\frac{1}{4}$ $k \leq m$ but since $\lfloor{\frac{k}{m}}\rfloor=1$ implies $k \geq m$ so $n = m^2+m$ Second Case : if $\lfloor{\frac{k}{m}}\rfloor = 2$ $k = 2m$ $2m+2 > 2 \sqrt{m^2+2m}$ $4m^2+8m+4 >4(m^2+2m)$ $4>0$ this is true, then $n = m^2+2m$ also satisfy so n that could be formed in term of $(m^2+m , m^2+2m)$ where m natural number is the solution of the equation Edit : I forgot to put sqrt