If two pairs of non-zero real numbers $\{a,b\},\{c,d\}$ have the property that the sum of the first pair equals the sum of the second pair, and the product of the first pair equals the product of the second pair, what can we conclude about the numbers?
I believe both pairs must be identical, but am struggling to prove this.
I've deleted my failed attempts to prove this from the OP; if you'd like to see them, they're in the history. Below are the successful proofs, along with how we reached them
Based on "ancient mathematician's" solution, I asked myself "How could I come up with that?" and worked out the following. I request feedback on how someone could come up with any of these solutions given that the direct approach above did not work.
Inspiration, Approach, Solution
Intuitively, we suspect that the pairs are equal, since if $c = ar$, then $b = a/r$, implying that the $\Delta a$ by multiplying $a$ by $r$ is negative the $\Delta b$ by dividing by $r$, which suggests $r = 1$. However, attempting to formalize this proves difficult. Why?
Because in fact it's not correct to conclude that $a = c$, only that $\{a,b\} =\{c,d\}$. It is the pairs that are equal, not the numbers. And equations are about numbers, not pairs!
How can we overcome this? One way is to ask "What object does respect pairs?" And the obvious answer is: A polynomial of degree 2. That is, if $f$ and $g$ are polynomials of degree 2 with identical pairs of roots, then $f = g$.
So we need to form a polynomial with roots $a,b$. And remembering "Math has a tendency to reward you if you respect its symmetries", we form a second polynomial $g$, aiming to show equality:
$$f(x) = (x-a)(x-b)\\ g(x) = (x-c)(x-d).$$
Now multiply out $$f(x) = x^2 - (a+b)x +ab \\ g(x) = x^2 - (c +d)x + cd$$ and the result is trivial.
Alternately...
A similar path, based on "user's" solution: We need to prove $a = c \lor a = d$, which can only be encoded via a 2nd degree polynomial: $(a-c)(a-d) = 0$. Working backwards temporarily: $$(a-c)(a-d) = 0\\ a^2 -a(c+d) +cd = 0\\ a^2 - a(a+b) + ab = 0\\ 0 = 0.$$ Now let's work forwards: $$a^2 - a(a+b) + ab = 0\\ a^2 -a(c+d) +cd = 0\\ (a-c)(a-d) = 0\\ a = c \lor a = d$$ as desired.