Suppose I have six positive numbers divided into two sets given by: $\{a, b, c\}$ and $\{d, e, f\}$. Now, I have relations as: $$a^2-b^2=d^2-e^2$$ $$b^2-c^2=e^2-f^2$$ $$a^2-c^2=d^2-f^2$$ I some feel that this is only possible when $a=d, b=e, c=f$, also it can be proved easily that the numbers will be identically ordered, i.e., if $a\geq b\geq c$ than $d \geq e \geq f$.
But is there some way to prove that $a=d, b=e, c=f$, One thing we may do is to show triangles created using side lengths $a,b,c$ and $d,e,f$ to be congruent. Is there a way to do so?
Nope. Consider $a^2=6$, $b^2=5$, $c^2=4$, $d^2=3$, $e^2=2$, $f^2=1$.