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

I have been asking the following question at MSE with an answer: $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ is true?

I found this relational expression by using computer. Then, I got interested in the generalization of this expression. After receiving an answer, I decided to start an easy example.

First, I got the following with a proof: $$\lfloor\sqrt n+\sqrt {n+1}\rfloor=\lfloor\sqrt {4n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor.$$

Note that $\lfloor x \rfloor$ is the largest integer not greater than $x$.

Proof: By the AM-GM inequality, we get $n\lt\sqrt{n(n+1)}\lt n+1/2$. Then, we get $$4n+1\lt\left(\sqrt n+\sqrt{n+1}\right)^2=4n+2-2\left(n+1/2-\sqrt{n(n+1)}\right)\lt4n+2.$$ By the way, there are no perfect squares in $\left(4n+1, 4n+2\right].$ (since $k^2\equiv0, 1(mod 4)$ for any natural number $k$, there is no natural number $k$ such that $k^2=4n+2$.) Now, the proof is completed.

Second, I've just got new relational expressions by using computer: $$\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}\rfloor=\lfloor\sqrt{9n+8}\rfloor,$$ $$\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\rfloor=\lfloor\sqrt{16n+20}\rfloor$$ for any natural numbre $n$.

I won't ask these expression because I'm going to prove these by myself.

Third, I haven't got any expression about 'six terms expression'. Then, I expect the following theorem would be proven true:

Theorem: For any real number $c$, there exists a natural number $n$ such that $$\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}+\sqrt{n+5}\rfloor\not=\lfloor\sqrt{36n+c}\rfloor.$$

Then, here is my question.

Question: The theorem I wrote is true? If it is true, please show me how to prove it. If it is not true, please give me a counterexample.

Any help would be appriciated.

• Have you been able to adapt the answer that you were given? If not, where did it break down? – Calvin Lin Aug 27 '13 at 15:35

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 $(k+1)^2$.