Four Integers

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 is half their 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