Generalization of $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$

Let $g(n)=\sqrt n+\sqrt{n+1}+\ldots+\sqrt{n+5} $ and $f(n)=\lfloor g(n)\rfloor$. When searching for a $c$ such that $$f(n) = \lfloor \sqrt{36n+c}\rfloor $$ for all $n$, each $n$ gives us some conditions on $c$, namely that $$f(n)\le\sqrt{36n+c}<f(n)+1 $$ that is $$\tag1f(n)^2-36n\le c<(f(n)+1)^2-36n.$$ This way, $n=1$ gives us $64\le c<85$ and $n=11$ gives us $88\le c<133$. Since these inequalites contradict each other, no $c$ works for all $n$, thus your theorem is true.

We can find suitable $c$ if we are allowed to ignore the first few values of $n$:

Note that for $k>0$ $$\sqrt{n+k}-\sqrt{n}=\frac{k}{\sqrt{n+k}+\sqrt n}\in\left(\frac{k}{2\sqrt{n+k}},\frac{k}{2\sqrt{n}}\right),$$ hence $$ 6\sqrt n+\frac{15}{2\sqrt{n+5}}<g(n) <6\sqrt n+\frac{15}{2\sqrt n}$$ and $$\tag236n+90\sqrt{\frac{n}{n+5}}+\frac{225}{4(n+5)}< g(n)^2<36n+90+\frac{225}{4n}.$$ Assume $n\ge20$, so that $4(n+5)\le 5n$. From $\sqrt{1+\frac 5n}<1+\frac{5}{2n}$ we have $\sqrt{\frac{n}{n+5}}>\frac1{1+\frac5{2n}}>1-\frac5{2n}$, so the left hand side of (2) becomes $$ g(n)^2>36n+90\left(1-\frac5{2n}\right)+\frac{45}{n}=36n+90-\frac{180}{n}.$$ Thus for $n>180$ we have $\sqrt{36n+89}<g(n)<\sqrt{36n+91}$ and hence $$ f(n)\in\bigl\{\lfloor\sqrt{36n+89} \rfloor,\lfloor\sqrt{36n+90} \rfloor,\lfloor\sqrt{36n+91} \rfloor\bigr\}.$$ Since $36n+90$ and $36n+91$ cannot be perfect squares ($x^2\equiv 18\pmod{36}$ and $x^2\equiv 19\pmod{36}$ have no solutions), we conclude $$ f(n)=\lfloor\sqrt{36n+89} \rfloor=\lfloor\sqrt{36n+90} \rfloor=\lfloor\sqrt{36n+91} \rfloor$$ at least for $n>180$, and by inspection for all $n>3$.

Remark: The above computations also suggest which value of $c$ we shold check in the general case: Similar as above we find that $$ \sqrt n+\ldots +\sqrt{n+k}\approx (k+1)\sqrt n+\frac{0+1+\ldots +k)}{2\sqrt n}=(k+1)\sqrt n+\frac{k(k+1)}{4\sqrt n}$$ and therefore $$ \left(\sqrt n+\ldots +\sqrt{n+k}\right)^2\approx (k+1)^2n+\frac{k(k+1)^2}{2}.$$ Further details depend on whether the integers near $\frac{k(k+1)^2}{2}$ are squares modulo $(k+1)^2$.