$\large y =\sqrt { x+20160 } -\sqrt { x-20160 }$

How many solutions exist for the above, such that both $x$ and $y$ are integers ?

The answer is 36.

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Relevant wiki: General Diophantine Equations - Problem SolvingSuppose that $m,x$ are integers such that $\sqrt{x + 20160} - \sqrt{x - 20160} = m$ . Then certainly $x,m \in \mathrm{N}$ , while $x \ge 20160$ and $m \le \sqrt{2 \times 20160} = 2\sqrt{10080}$ . Thus, on squaring, $\begin{array}{rcl} 2x - 2\sqrt{x^2 - 20160^2} & = & m^2 \\ 2x - m^2 & = & 2\sqrt{x^2 - 20160^2} \\ (2x - m^2)^2 & = & 4\big[x^2 - 20160^2\big] \\ 4m^2x & = & m^4 + 4\times20160^2 \end{array}$ Hence $m^4$ must be even, and so $m$ must be even. Thus $m = 2n$ for some $n \in \mathrm{N}$ , and $n^2x \; = \; n^4 + 10080^2$ Thus $n$ must divide $10080$ , and hence we must have $x \; = \; n^2 + \big(\tfrac{10080}{n}\big)^2 \qquad \qquad m \; = \; 2n$ for any positive integer divisor $n$ of $10080$ for which $n \le \sqrt{10080}$ . Since $10080 = 2^5 \times 3^2 \times 5 \times 7$ , there are $6\times3\times2\times2 = 72$ positive integer divisors of $10080$ . Since $10080$ is not a perfect square, exactly half of these are less than or equal to $\sqrt{10080}$ . Thus there are $\tfrac{72}{2} = \boxed{36}$ solutions in integers to this equation.