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Prove that: $\sqrt{2-\frac{1}{1^2+\sqrt{1^4+\frac{1}{4}}}}+\sqrt{2-\frac{1}{2^2+\sqrt{2^4+\frac{1}{4}}}}+\sqrt{2-\frac{1}{3^2+\sqrt{3^4+\frac{1}{4}}}}+\cdots+\sqrt{2-\frac{1}{119^2+\sqrt{119^4+\frac{1}{4}}}}$ is an integer.

I noticed that $n^4+\frac{1}{4}$ is rather interesting, because it would be written as $(n^2+n+\frac{1}{2})(n^2-n+\frac{1}{2})$, but I do not know how to connect this to the rest of the problem. The square roots seem to further complicate the problem. Furthermore, I don't see what's special with 119, but upon using a program, I noticed that the only such numbers (such that the sum is an integer) below 500 are 3, 20, and 119.

The Math Troll
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  • Often in problems of this nature, consider the sum of these numbers taken pairwise with the pairs being $n, 120-n$. This may not work but is worth a try. – Yunus Syed Mar 31 '17 at 03:55
  • I don't think that this would work, because as $n$ goes large, the value of the fraction becomes closer and close to $\sqrt{2}$. – The Math Troll Mar 31 '17 at 04:04
  • How do you know it's actually an integer, rather than something with a lot of zeros after the decimal point? – mathworker21 Mar 31 '17 at 04:20

1 Answers1

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Since $$\frac{1}{k^2+\sqrt{k^4+\frac 14}}=\frac{k^2-\sqrt{k^4+\frac 14}}{(k^2)^2-(k^4+\frac 14)}=-4k^2+2\sqrt{4k^4+1}$$ we have $$\sum_{k=1}^{n}\sqrt{2-\frac{1}{k^2+\sqrt{k^4+\frac 14}}}=\sum_{k=1}^{n}\sqrt{2+4k^2-2\sqrt{4k^4+1}}\tag1$$

Here, since we have that $$4k^4+1=4k^4+4k^2+1-4k^2=(2k^2+1)^2-(2k)^2=(2k^2+2k+1)(2k^2-2k+1)$$ and that $$(2k^2+2k+1)+(2k^2-2k+1)=2+4k^2$$ we have $$\begin{align}(1)&=\sum_{k=1}^{n}\sqrt{\left(\sqrt{2k^2+2k+1}-\sqrt{2k^2-2k+1}\right)^2}\\\\&=\sum_{k=1}^{n}\left(\sqrt{2k^2+2k+1}-\sqrt{2k^2-2k+1}\right)\\\\&=\sum_{k=1}^{n}\left(\sqrt{2k^2+2k+1}-\sqrt{2(k-1)^2+2(k-1)+1}\right)\\\\&=\sqrt{2n^2+2n+1}-\sqrt{1}\end{align}$$

This is an integer for $n=119$ since $2\cdot 119^2+2\cdot 119+1=169^2$.

Added :

$(1)$ is an integer if and only if there is an integer $m$ such that $$2n^2+2n+1=m^2\iff (2n+1)^2-2m^2=-1$$ This is a Pell's equation, so we have $$n=-\frac 12+\left(\frac{1-\sqrt 2}{4}\right)\cdot (3-2\sqrt 2)^k+\left(\frac{1+\sqrt 2}{4}\right)\cdot (3+2\sqrt 2)^k$$ where $k=1,2,3,\cdots$.

These satisfy $a_n=6a_{n-1}-a_{n-2}+2$ with $a_1=3,a_2=20$ : $$a_3=119,\quad a_4=696,\quad a_5=4059,\quad a_6=23660,\quad \cdots$$

mathlove
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