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  1. Let m = 1 $$\sum_{k=0}^m \sqrt{n+k} = \sqrt{n} + \sqrt{n+1}$$

$$\sqrt{n} + \sqrt{n+1} = \alpha \in \mathbb{Q}$$ $$(\alpha - \sqrt{n})^2 = (\sqrt{n + 1})^2 \Leftrightarrow \alpha^2 - 2\alpha\sqrt{n} + n = n + 1 \Rightarrow \sqrt{n} = \frac{\alpha^2 - 1}{2\alpha} \in \mathbb{Q} \Rightarrow n = r^2, \; r \in \mathbb{N}$$ $$(\alpha - \sqrt{n+1})^2 = (\sqrt{n})^2 \Leftrightarrow \alpha^2 - 2\alpha\sqrt{n+1} + n + 1 = n \Rightarrow \sqrt{n + 1} = \frac{\alpha^2 + 1}{2\alpha} \in \mathbb{Q} \Rightarrow n = t^2 - 1, \; t \in \mathbb{N}$$

$$t^2 - r^2 = 1 \Leftrightarrow (t - r)(t + r) = 1 \Leftrightarrow t - r = \pm 1, t + r = \pm 1 \Rightarrow r = 0 \Leftrightarrow n = 0 \Rightarrow \alpha \not \in \mathbb{Q}$$

  1. Let m = 2 $$\sum_{k=0}^m \sqrt{n+k} = \sqrt{n} + \sqrt{n+1} + \sqrt{n+2}$$

$$\sqrt{n} + \sqrt{n+1} + \sqrt{n+2} = \alpha \in \mathbb{Q}$$ $$(\alpha - \sqrt{n+2})^2 = (\sqrt{n} + \sqrt{n+1})^2 \Leftrightarrow \alpha^2 - 2\alpha\sqrt{n+2} + n + 2 = 2n + 1 + 2\sqrt{n(n+1)}, \; \beta = \frac{\alpha^2 - n + 1}{2},\; \beta \in \mathbb{Q}$$ $$(\beta - \alpha\sqrt{n+2})^2 = n(n+1) \Leftrightarrow \beta^2 - 2\alpha\beta\sqrt{n+2} + n + 2 = n(n+1) \Rightarrow \sqrt{n + 2} \in \mathbb{Q} \Rightarrow n = t^2 - 2, \; t \in \mathbb{N}$$

$$(\alpha - \sqrt{n+1})^2 = (\sqrt{n} + \sqrt{n+2})^2 \Leftrightarrow \alpha^2 - 2\alpha\sqrt{n+1} + n + 1 = 2n + 2 + 2\sqrt{n(n+2)}, \; \gamma = \frac{\alpha^2 - n - 1}{2}, \; \gamma \in \mathbb{Q}$$ $$(\gamma - \alpha\sqrt{n+1})^2 = n(n+2) \Leftrightarrow \gamma^2 - 2\alpha\gamma\sqrt{n+1} + n + 1 = n(n+2) \Rightarrow \sqrt{n + 1} \in \mathbb{Q} \Rightarrow n = r^2 - 1, \; r \in \mathbb{N}$$ $$t^2 - r^2 = 1 \Leftrightarrow (t - r)(t + r) = 1 \Leftrightarrow t - r = \pm 1, t + r = \pm 1 \Rightarrow r = 0 \Leftrightarrow n = -1 \Rightarrow \alpha \not \in \mathbb{Q}$$

So, for m = 1, 2 this is true.

flamble
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    Some commentary would be appreciated. – Shaun Apr 17 '23 at 14:42
  • @Shaun I solved the problem for the case m = 1, 2, it is not clear what can be done in the general case – flamble Apr 17 '23 at 14:48
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    That's better. Thank you, but what I meant was that the question would benefit from some English, instead of it just being a wall of mathematical symbols; explain your reasoning in words with an [edit], please. – Shaun Apr 17 '23 at 14:55
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    This is certainly true. Here's a link to a more general statement, also true: https://mathoverflow.net/questions/396205/sum-of-square-roots-of-natural-numbers . And another: https://math.stackexchange.com/questions/30687/the-square-roots-of-different-primes-are-linearly-independent-over-the-field-of . – mjqxxxx Apr 17 '23 at 22:37

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