HMMT-Feb 2014 alg, Problem 2

Find the integer closest to \[ \frac{1}{\sqrt[4]{5^{4}+1}-\sqrt[4]{5^{4}-1}} \]

Solution: Let $x=\left(5^{4}+1\right)^{1 / 4}$ and $y=\left(5^{4}-1\right)^{1 / 4}$. Note that $x$ and $y$ are both approximately 5. We have \[ \begin{aligned} \frac{1}{x-y} & =\frac{(x+y)\left(x^{2}+y^{2}\right)}{(x-y)(x+y)\left(x^{2}+y^{2}\right)}=\frac{(x+y)\left(x^{2}+y^{2}\right)}{x^{4}-y^{4}} \\ & =\frac{(x+y)\left(x^{2}+y^{2}\right)}{2} \approx \frac{(5+5)\left(5^{2}+5^{2}\right)}{2}=250 \end{aligned} \] Note: To justify the $\approx$, note that $1=x^{4}-5^{4}$ implies \[ 0<x-5=\frac{1}{(x+5)\left(x^{2}+5^{2}\right)}<\frac{1}{(5+5)\left(5^{2}+5^{2}\right)}=\frac{1}{500} \] and similarly $1=5^{4}-y^{4}$ implies \[ 0<5-y=\frac{1}{(5+y)\left(5^{2}+y^{2}\right)}<\frac{1}{(4+4)\left(4^{2}+4^{2}\right)}=\frac{1}{256} \] Similarly, \[ 0<x^{2}-5^{2}=\frac{1}{x^{2}+5^{2}}<\frac{1}{2 \cdot 5^{2}}=\frac{1}{50} \] and \[ 0<5^{2}-y^{2}=\frac{1}{5^{2}+y^{2}}<\frac{1}{5^{2}+4.5^{2}}<\frac{1}{45} \] Now \[ |x+y-10|=|(x-5)-(5-y)|<\max (|x-5|,|5-y|)<\frac{1}{256} \] and similarly $\left|x^{2}+y^{2}-2 \cdot 5^{2}\right|<\frac{1}{45}$. It's easy to check that $(10-1 / 256)(50-1 / 45)>499.5$ and $(10+1 / 256)(50+1 / 45)<500.5$, so we're done. $\fbox{250}$.

Answer: 250