I'm trying to find all $(a,x,b,y) \in \mathbb{R}^4_{\geq 0}$ satisfying:
$a^2+x^2=b^2+y^2=ax+by$
It is clear that $(t,t,t,t)$ is a solution of given equations for any non negative real number $t$ . Also, if any two numbers from $a,x,b$ and $y$ are equal then all four will be equal. Is there any other solution of given equations?
We have that $$a^2+x^2+b^2+y^2 = 2ax + 2by$$
Rearranging yields $$a^2-2ax+x^2+b^2-2by+y^2 = (a-x)^2 + (b-y)^2 = 0$$
Since squares are non-negative, we must have that $a=x$ and $b=y$.
Thus, we have that $2a^2 = 2b^2$ and thus $a=b$.
This gives us all solutions to be of the form $(t,t,t,t)$.
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If we take the restriction of $\mathbb{R}^+_0$ off and as only $\mathbb{R}$, we have $(t,t,\pm t,\pm t)$ where the $\pm$ are dependent of each other.