Notice that if $ \alpha <0$ then it is not true for $ n=1$ $ [\sqrt {1} + \sqrt {1 + \alpha}] < 2=[\sqrt {5}]$ Same for $ 3 \le \alpha$ $ [\sqrt {1} + \sqrt {1 + \alpha}]=[1 + \sqrt {1 + \alpha}] >2=[\sqrt {5}]$ $ 1 \le \alpha$ $ 1/2 \le \alpha /2$ $ n^2 < n^2 + n \alpha$ $ n < \sqrt {n^2+n \alpha}$ $ n+1/2 \le \alpha /2+ \sqrt {n^2+n \alpha}$ $ 2n+1 \le \alpha +2 \sqrt {n^2+n \alpha}$ $ 4n+1 \le 2n+\alpha+ 2 \sqrt {n^2+n \alpha}$ $ \sqrt {4n + 1} \le \sqrt {n} + \sqrt {n + \alpha}$ $ 1\le \alpha < 3$ $ \alpha /2< 3/2$ $ n^2+n \alpha < n^2 +3 n < n^2$ $ \sqrt {n^2+n \alpha} < n$ $ \alpha /2+\sqrt {n^2+n \alpha}< 3/2+n$ $ \alpha /2+\sqrt {n^2+n \alpha}< n+1+\sqrt {4n + 1}$ $ \alpha +2 \sqrt {n^2+n \alpha}< 2n+2+2 \sqrt {4n + 1}$ $ 2n+\alpha+ 2 \sqrt {n^2+n \alpha}< 4n+1+2 \sqrt {4n + 1}+1$ $ (\sqrt {n} + \sqrt {n + \alpha})^2 < (\sqrt {4n + 1}+ 1)^2$ $ \sqrt {n} + \sqrt {n + \alpha}- \sqrt {4n + 1} < 1$ then $ 0 \le \sqrt {n} + \sqrt {n + \alpha}- \sqrt {4n + 1} < 1$ for $ 1\le \alpha <3$ Idem for $ 0\le \alpha \le 1$ but $ 0 \le \sqrt {4n + 1}-\sqrt {n} + \sqrt {n + \alpha}< 1$

Peter wrote: Find all real numbers $ \alpha$ for which the equality \[ \lfloor \sqrt {n} + \sqrt {n + \alpha}\rfloor = \lfloor \sqrt {4n + 1}\rfloor \] holds for all positive integers $ n$. Because $ (\sqrt x )'' < 0$ we get $ \sqrt x + \sqrt {x + \alpha } < 2\sqrt {x + \alpha/2} = \sqrt {4x + 2\alpha}$. If $ \alpha>2$ for $ n=k^2-1$ we get $ \sqrt {k^2 - 1} + \sqrt {k^2 - 1 + \alpha } < 2k$. It is equavalent to $ (\alpha -2)k^2 < \alpha ^2$ and false for big $ k$. Therefore solution is $ 9 - 6\sqrt 2 \le \alpha \le 2$. If it must be true for $ n = 0$ solution is $ 1\le \alpha \le 2.$

Rust: Do you have a counterexample for $ \alpha >2$ ? Could you please find the mistake in my previous message?

If $ \alpha > 2$, then for $ n = k^2 - 1, \ k > \frac {\alpha }{\sqrt {\alpha -2}}$ give contradition.

I agree with Rust that the equality holds for all $n$ if, and only if, $9 - 6\sqrt{2} \le \alpha \le 2$. And note that $9 - 6\sqrt{2} \approx 0.51$. Proof. For $n = 1$ we can readily check that the equality holds for all $\alpha$ with $0 \le \alpha < 3$, so in particular for $\alpha \in [9 - 6\sqrt{2}, 2]$. We may therefore assume $n \ge 2$. To prove our claim, we must show both the necessity and the sufficiency of both the lower and upper bound. To prove the necessity of the lower bound, we look at $n=2$. It follows that $\alpha$ must be such that $\sqrt{2+\alpha} \ge 3 - \sqrt{2}$. Squaring both sides and subtracting $2$ gives $\alpha \ge 9 - 6\sqrt{2}$. To prove the necessity of the upper bound, assume by contradiction $\alpha = 2 + \epsilon$ with $\epsilon > 0$ and choose $n > \frac{1}{\epsilon}$ such that $n+1$ is a square, say $n+1 = k^2$. Then we have the following inequalities: \begin{align*} 4n(n + 2 + \epsilon) &> 4n^2 + 8n + 4 \\ 2\sqrt{n}\sqrt{n+2+\epsilon} &> 2n + 2 > 2n + 2 - \epsilon \\ 2n+2+\epsilon + 2\sqrt{n}\sqrt{n+2+\epsilon} &> 4n + 4 \\ (\sqrt{n} + \sqrt{n+2+\epsilon})^2 &> 4n + 4 \\ \sqrt{n} + \sqrt{n+2+\epsilon} &> 2\sqrt{n+1} = 2k \end{align*} On the other hand, $\sqrt{4n+1} < \sqrt{4n+4} = 2k$, contradicting the equality. To prove that these bounds are also sufficient, first we claim that $\sqrt{n} + \sqrt{n+\alpha} \ge \sqrt{4n+1}$ holds for all $n \ge 2$, whenever $\alpha \ge 9 - 6\sqrt{2}$. It suffices to prove this for $\alpha = 9 - 6\sqrt{2}$, so let us assume that for the moment. By the quadratic formula we see $8(2\alpha-1) - (\alpha-1)^2 = -\alpha^2 + 18\alpha - 9 = 0$. For $n \ge 2$ we therefore have the following inequalities: \begin{align*} 4n(2\alpha-1) - (\alpha-1)^2 &\ge 0 \\ 8\alpha n &\ge 4n + 1 - 2\alpha + \alpha^2 \\ 4n^2 + 4\alpha n &\ge 4n^2 + 4n + 1 - 4\alpha n - 2\alpha + \alpha^2 \\ 4n(n + \alpha) &\ge (2n+1)^2 - 2(2n+1)\alpha + \alpha^2 = ((2n+1) - \alpha)^2 \\ 2\sqrt{n}\sqrt{n+\alpha} &\ge 2n + 1 - \alpha \\ 2n + \alpha + 2\sqrt{n}\sqrt{n+\alpha} &\ge 4n + 1 \\ (\sqrt{n} + \sqrt{n+\alpha})^2 &\ge 4n + 1 \\ \sqrt{n} + \sqrt{n+\alpha} &\ge \sqrt{4n+1} \end{align*} With the above inequality in mind, we can finish the proof by showing that $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le \lfloor \sqrt{4n+1} \rfloor$ holds for all $n$. Assume $k^2 \le n < (k+1)^2$. We then have to distinguish between $n < k^2 + k$ or $n \ge k^2+k$. If $k^2 \le n \le k^2 + k - 1$, then $\lfloor \sqrt{4n+1} \rfloor = 2k$ and $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le \lfloor 2\sqrt{n+1} \rfloor = 2k$, since $\sqrt{n+1} < k + \frac{1}{2}$. On the other hand, if $k^2 + k \le n \le k^2 + 2k$, then $\lfloor \sqrt{4n+1} \rfloor = 2k+1$. And $\lfloor \sqrt{n} + \sqrt{n + 2} \rfloor \le 2k+1$, since $\sqrt{n} + \sqrt{n + 2} < 2\sqrt{n+1} \le 2k+2$.