# An Application of Quadratic Formula

Here's a curious fact:

$\displaystyle \bigg(\frac{\sqrt{2+\sqrt{1}}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{1}}+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}}-1\bigg)^2=2.$

It would be a risky venture to prove the unappealing identity directly. Interestingly, it is only a special case of a more general equality that motivated the authors of a problem from the 2009 Japanese Mathematical Olympiad [Lecture Notes, p 30]:

Evaluate

$\displaystyle \frac{\displaystyle\sum_{n=1}^{99}(10+\sqrt{n})}{\displaystyle\sum_{n=1}^{99}(10-\sqrt{n})}$

As a matter of fact, the following identity holds:

For $N\gt 1$,

$\displaystyle \frac{\displaystyle\sum_{n=1}^{N^{2}-1}(N+\sqrt{n})}{\displaystyle\sum_{n=1}^{N^{2}-1}(N-\sqrt{n})} = 1+\sqrt{2}.$

The identity at the top is obtained by letting $N=2$. I have a feeling that the more general formulation may be more suggestive of a possible solution. The choice of a specific value for $N$, be it $10$ or $2$ seems (to me at least) to make the problem more obscure. Of course it is wonderful that the expression is independent of $N$.

Solution

### References

1. Xu Jiagu, Lecture Notes on Mathematical Olympiad Courses, v 8, (For senior section, v 2), World Scientific, 2012

For $N\gt 1$,

$\displaystyle \frac{\displaystyle\sum_{n=1}^{N^{2}-1}(N+\sqrt{n})}{\displaystyle\sum_{n=1}^{N^{2}-1}(N-\sqrt{n})} = 1+\sqrt{2}.$

The starting point for a solution is the following observation:, for positive $a$ and $b$,

$\sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}.$

In that formula, $a$ and $b$ can be sought of as the roots of the quadratic equation

$x^{2}-(a+b)x+ab = 0.$

Such a view will prove helpful, provided we can find $a$ and $b$ that satisfy Viète's formulas, $a+b=2N$ and $ab=n$. So we check the quadratic equation

$x^{2}-2Nx+n = 0.$

The discriminant of the equation equals $4(N^{2}-n)$ and is positive exactly when $1\le n \lt N^2$. The two roots are $N\pm\sqrt{N^{2}-n}$. Taking $a=N+\sqrt{N^{2}-n}$ and $b=N-\sqrt{N^{2}-n}$ we see that,

$\sqrt{N+\sqrt{N^{2}-n}}+\sqrt{N-\sqrt{N^{2}-n}} = \sqrt{2N+2\sqrt{n}}.$

Thus we define, say, $\displaystyle S=\sum_{n=1}^{N^{2}-1}(N+\sqrt{n})$ and $\displaystyle T=\sum_{n=1}^{N^{2}-1}(N-\sqrt{n})$. What we just found shows that

\displaystyle \begin{align} \sqrt{2}S &= \sum_{n=1}^{N^{2}-1}(\sqrt{N+\sqrt{N^{2}-n}}+\sqrt{N-\sqrt{N^{2}-n}}) \\ &= \sum_{n=N^{2}-1}^{1}(\sqrt{N+\sqrt{N^{2}-n}}+\sqrt{N-\sqrt{N^{2}-n}}) \\ &= \sum_{n=1}^{N^{2}-1}(\sqrt{N+\sqrt{n}}+\sqrt{N-\sqrt{n}}) \\ &= S + T, \end{align}

which implies that $\displaystyle\frac{S}{T}=\frac{1}{\sqrt{2}-1}=1+\sqrt{2}$.

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