The answer is 65.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Well, I enjoyed solving this! First, begin by noticing something. The number $n$ that we are looking for is the average of two squares. We can see this by observing that $\sqrt{n-x_1}$ and $\sqrt{n+x_1}$ are both integral. We can further observe that it can be written as the average of squares in two different ways! So let's find $2n$ , which will be the

sum(remember, their average ishalftheir sum for two numbers) of two squares in two different ways. This is a really fascinating branch of Number Theory! I read this article that I thought was supremely interesting. You'll note that he even lists some of these "semi"-taxicab numbers. But we have to be careful: the sum of the squares must be even for $n$ to be integral. From our knowledge of parity, either all four squares are odd, or they're all even. Thus, the smallest of the numbers that fits these requirements is $130=3^2+11^2=7^2+9^2$ and we want $\frac{130}{2}=\boxed{65}$ . Another thing I tried doing is looking for Pythagorean triples that had the same hypotenuse. The smallest one of these numbers is $65^2=16^2+63^2=33^2+56^2$ . This wouldn't work though, because $65^2$ is odd. Anyways, fascinating problem @Daniel Liu ! :D