Find a diagonal form of the quadratic form $f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j$ Find a diagonal form of the quadratic form
$$f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j.$$ 
It turned out to be such a problem:
How to change quadratic form $f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j$ into diagonal form?
We can solve it with congruent transformation, although it is a little complex.
Besides, I tried to transform it into a diagonal form by orthogonal transformation, since its quadratic matrix has the same diagonal elements and the same non-diagonal elements(I thought that I can find a simple way to find its eigenvalues and eigenvectors, but failed. Maybe you can try it).
Is there any other way?

This is too long for the comment box.
The essence of your exercise is given in the following example.
Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$\begin{align*}
f(x, y, z) &= (x^2 + xy + xz) + y^2 + z^2 + yz \\
&= \left( x + \dfrac{y}{2} + \dfrac{z}{2} \right)^2 - \frac{y^2}{4} - \frac{z^2}{4} -\frac{yz}{2} + y^2 + z^2 + yz\\
&=u^2 + \dfrac{3y^2}{4} + \dfrac{yz}{2} + \dfrac{3z^2}{4}
\end{align*},$$
where $u = x + \dfrac{y}{2} + \dfrac{z}{2}.$ Now, we can get rid of the second variable, $y$ in my case, by, again, completing the square
$$\begin{align*}
f(x, y, z) &= u^2 + \left( \dfrac{\sqrt{3}y}{2} + \frac{z}{2\sqrt{3}} \right)^2-\dfrac{z^2}{12}+\dfrac{3z^2}{4} = u^2 + v^2+\dfrac{5}{6}z^2
\end{align*},$$
which is a required representation of $f.$
The case with $x_1, \ldots, x_p$ is just guessing a pattern and using induction (I am guessing).
