How would I solve $x^2-4x=y^2-4y$ without knowing the answer beforehand?

I take the opportunity of this question to consider a more general issue.

Here is a methodology that can be applied to any equation of the form :

$$f(x_1)=f(x_2) tag{1}$$

where $f$ is a function from $I$ (subset of $mathbb{R}$) to $J:=f(I)$.

(I intentionally take variables $(x_1,x_2)$ instead of $(x,y)$)

Of course, $(x,x), x in I$ is a natural (trivial) set of solutions.

Once the study of variations of $f$ has been done, operate a segmentation of the range $J$ of $f$ into the following components : $$J=J_1 cup J_2 cup J_3 cup cdots$$

where $J_k$ is defined as the subset of $J$ of all $y$ that $f(x)=y$ has $k$ distinct solutions. As a consequence, each $J_k$ will generate $k(k-1)$ non-trivial solutions to equation (1) ; for example, when $k=3$, we will have solutions :

$$(x_1,x_2), (x_1,x_3), (x_2,x_3) text{and, symmetrically} (x_2,x_1), (x_3,x_1), (x_3,x_2)$$

Remark : the "trivial" cases correspond to $J_1$.

Let us consider the following example :

$$y=f(x)=x(x-1)(x-2)(x-3) text{on} I=[0, +infty) text{and} J=[-1,+infty).$$

with the following curve, possessing in particular a local maximum in $(-tfrac12,a:=tfrac{9}{16})$

We will have

• $J_1=(a,+infty)$ generating "trivial solutions" to (1),

• $J_2={0,a}$ (set with two elements), generating sets of two non-trivial solutions,

• $J_3=(0,a)$ (a whole line segment !), generating sets of $3 times 2$ solutions,

• $J_4=(-1,0)$ generating sets of $6 times 2$ solutions.

Now, let us return to the case of function $y=f(x)=x^2-4x$ (see Fig. 2) :

we will clearly have :

• $J_1={y=-4}$ corresponding to point $x=2$ giving trivial solution $(x_1,x_1)=(2,2)$

• $J_2=(-4,+infty)$, yielding 2 solutions $(x_1,x_2)$ and $(x_2,x_1)$ obtained by solving, for each $y in (-4,+infty)$ the quadratic equation :

$$x^2-4x=y$$

(take care, once again : this $y$ has nothing to do with the $y$ of the initial question $f(x)=f(y)$ !)

giving the two solutions :

$$x_{1,2}=2 pm sqrt{4+y^2}tag{2}$$

A first remark is that

$$x_1+x_2=4 text{which is the solution "as presented"}. tag{3}$$

A second and final remark is that (3) describes exactly all solutions because $x_1$ (and the same for $x_2$) can take any value : for every value of $x_1$, one can find a value of $y$ such that (2) is fulfilled.