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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.

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  • $\begingroup$ Your question has become very difficult to read (in terms of notation) due to cutting wrong part in edits. Answers also don't really make sense. You should really consider at least restating in the post, what are a,b,c,d and which property they satisfy. $\endgroup$
    – Surb
    Dec 14, 2023 at 8:41
  • $\begingroup$ @Surb I added in explicit $\{a,b\},\{c,d\}$ (although I'm not convinced it's necessary). Is there anything else that would help you? If yes, please let me know. If not, you can delete your comment now. $\endgroup$ Dec 15, 2023 at 2:45

13 Answers 13

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We have $a$ and $b$ are the roots of $X^2-(a+b)X+ab$.

We have $c$ and $d$ are the roots of $X^2-(c+d)X+cd$.

But these quadratics are equal.

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    $\begingroup$ Exactly, Vieta's formulas! +1 $\endgroup$
    – Vasili
    Dec 12, 2023 at 15:12
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    $\begingroup$ But how did you pull out of a hat "Let's make a polynomial $(X-a)(X-b)$ and compare it to $(X-c)(X-d)$"? What insight led you to take this step? $\endgroup$ Dec 12, 2023 at 15:31
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    $\begingroup$ It comes from recognising an application of Vieta's formulas. If you're familiar with them, then "this is the sum of two numbers" and "this is the product of two numbers" very quickly translates to "these are the coefficients of a quadratic". $\endgroup$
    – ConMan
    Dec 12, 2023 at 23:15
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    $\begingroup$ @ConMan No doubt it ought to have come that way. But in fact I had never heard the name "Vieta's formulas" until I was about 74 years old, some 60 years after the facts about quadratics had been hard-wired into my head! $\endgroup$ Dec 13, 2023 at 8:01
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    $\begingroup$ @SRobertJames Here’s how you might derive it without knowing Vieta’s formulas in advance. Approach it from the question of: Can we solve for $a$ and $b$, given their sum $s = a+b$ and product $p=ab$? Solving the latter equation for $b$ gives $b = p/a$. Then we can substitute it into the former equation, giving $s = a + p/a$; now rearranging to a standard quadratic gives $a^2 - sa + p = 0$. Either by noticing it’s symmetric between $a$ and $b$, or simply by repeating the same process with $b$ as your goal, you find that $b$ satisfies the same quadratic, $b^2 - sb + p = 0$. $\endgroup$ Dec 14, 2023 at 15:03
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We can conclude that $\{c,d\}=\{a,b\}$.

Indeed substituting $d=a+b-c$ into the $cd=ab$ one obtains for $c$ the equation: $$ 0=c^2-(a+b)c+ab=(c-a)(c-b). $$

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  • $\begingroup$ I like this. The inspiration seems to be "We can't conclude that $c=a$ or $c=b$, or relate $c$ to either one alone. We can only relate $c$ to both, that $c = a \lor c =b$. So let's do that by writing $c$ in terms of $a$ and $b$, eliminating $d$, and by recognizing the degree 2 polynomial, we have our OR statement. $\endgroup$ Dec 12, 2023 at 15:53
  • $\begingroup$ @SRobertJames Yes, it is correct. Without the condition $c\le d$ you have a free choice. Of course the latter relation assumes that both $a$ and $b$ are real. But the equation is valid quite generally for complex numbers as well. $\endgroup$
    – user
    Dec 12, 2023 at 16:14
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    $\begingroup$ Isn't $(c, d)$ is an ordered pair, such that $(1, 2)\neq (2, 1)$? Did you mean $\lbrace c, d\rbrace = \lbrace{a, b\rbrace}$? $\endgroup$
    – Kaz
    Dec 15, 2023 at 0:01
  • $\begingroup$ @Kaz In the initial version of the question, which I answered, there was an additional condition $a\le b$, $c\le d $. $\endgroup$
    – user
    Dec 16, 2023 at 7:55
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Here's another rather different solution. We know that the pairs have the same sum. Observe that they must be identical if they have the same difference too. Indeed, $|a-b|^2$ $= (a-b)^2$ $= (a+b)^2-4·a·b$ $= (c+d)^2-4·c·d$ $= (c-d)^2$ $= |c-d|^2$, and so $|a-b| = |c-d|$.

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Here's a "visual" proof you can do in your head without having to actually write down anything:

The question is obviously equivalent to the following:

Q: Are there two noncongruent rectangles with the same circumference and the same area?

Any rectangle with side lengths $a$ and $b$ can be represented by the points $(a,b)$ and $(b,a)\in\mathbb{R}_+^2$. These rectangles' iso-circumference sets are lines, and their iso-area sets are hyperbolas. A line and a hyperbola have at most two intersections, so the answer to Q is No.

enter image description here

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  • $\begingroup$ As a first-time contributor I'd be happy if one of those users who suggested "Delete" for my answer would provide an explanation of what is wrong with it. $\endgroup$
    – VARulle
    Dec 19, 2023 at 8:50
  • $\begingroup$ I certainly wouldn't have suggested "Delete" for this answer. If you'd like to know how it might be improved, you might explain "rectangles iso-circumference sets are lines, and their iso-area sets are hyperbolas," which is probably not taken for granted by most people on this page. I'd also explain how at most two intersections implies congruence of rectangles implies pairs are identical. $\endgroup$ Dec 27, 2023 at 17:42
  • $\begingroup$ @SRobertJames, O.k., maybe this is self-explaining only to economists. They deal with these iso-something sets a lot... $\endgroup$
    – VARulle
    Dec 29, 2023 at 11:45
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Let $s = a+b = c+d$, so $b = s-a$ and $d=s-c$. It follows that $(a-c)s = a^2-c^2 = ((a+c)(a-c)$ and a=c, b=d or a=d, b=c.

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Another way to see it is as follows. Fix the sum $S$. Let $x$ be the larger number in a pair with sum $S$; clearly we have $x\in[S/2,\infty)$. Now the product is $x(S-x)=xS-x^2$. This has derivative $S-2x<0$ for $x\in(S/2,\infty)$, so it is strictly decreasing. Therefore for any target product $P$, there is at most one value of $x\in[S/2,\infty)$ where the product equals $P$.

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  • $\begingroup$ Interesting. What motivated you towards this approach? $\endgroup$ Dec 13, 2023 at 14:10
  • $\begingroup$ @SRobertJames the AM-GM inequality. $\endgroup$ Dec 13, 2023 at 20:00
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Let $c = ar$. Then $d = b/r$, giving $$a + b = ar + b/r.$$ Now $r = 1$ is clearly a solution. Other solutions for $r$ obviously exist; however, they only seem to result in the trivial alternate solution $a = d, b = c$ (which I excluded by assuming $a \leq b, c \leq d$). I do not know how to prove this last step.

You were on the right track.

You have an equation in $r$ that involves $r$, $1/r$ and a constant term. This is equivalent to a quadratic equation in $r$; just multiply both sides of the equation with $r$ to get this quadratic equation:

$$ar^2 -(a+b)r + b = 0$$

Solving this quadratic equation gives two real solutions, $r = 1$ and $r = b/a$, with equality iff the discriminant is 0, which happens when $a=b$.

If $r = 1$ then $(c, d) = (a, b)$. If $r = b/a$ then $(c, d) = (b, a)$.

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  • $\begingroup$ Aha! My mistake was to think "I must prove $r = 1$". In truth, $r = b/a$ proves the equality of the pairs just as well. It seems it always comes back to the central challenge of this problem: We need to prove equality of pairs, not numbers, and that requires thinking in quadratics (not individual variables one at a time). $\endgroup$ Dec 13, 2023 at 14:09
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    $\begingroup$ @SRobertJames I have a different approach: I didn't think "I must prove r=1" nor "I must prove r =b/a" nor "I must prove equality of pairs". I just thought "We want to find the possible values of r, which means we must solve the equation $a+b = ar + b/r$, and I happen to know how to solve this equation". $\endgroup$
    – Stef
    Dec 13, 2023 at 14:11
  • $\begingroup$ It's a relatively well-known trick that you should probably remember: if you ever encounter an equation like $ax + b + c/x = 0$ again, it's equivalent to $ax^2 + bx + c = 0$. $\endgroup$
    – Stef
    Dec 13, 2023 at 14:13
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In your question, you mention, “two pairs of non-zero real numbers”.

The equality of the two pairs is true even if you allow zeroes:

Suppose one pair contains at least one zero.
Then this pair is of the form $(0, A)$.
Its sum is $A$ and its product is $0$.

Therefore the other pair has a sum of $A$ and a product of $0$.
These conditions imply the other pair is also $(0, A)$.

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Here's a picture of the answers based on quadratic polynomials. Fix $S = a+b = c+d$ as the sum. Then for a fixed product $ab = cd = P = x(S-x)$, there are two solutions for $x$ (including multiplicity, see How can we prove that a quadratic equation has at most 2 roots? for $x(S-x)-P$). Conclude $x$ must be one of $a, b$ (or both if equal) and one of $c,d$.

In this picture we fix $P = 0.1$ (the green line) and $S = 1$ (determines the shape of the parabola). We can get $S = 1$ WLOG by scaling $a,b,c,d$ by $1/S$.

image of parbola

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  • $\begingroup$ Nice. I read this as: Given that the product is fixed, if we represent it as $x(S-x)$, we have only two choices for $x$; call them $\alpha$ and $\beta$. But that implies if $x = \alpha$ then $S-x = \beta$ and vice versa - so the pairs must be congruent. Is that correct? What do you mean by $a',b',c',d'$? I don't see those defined. $\endgroup$ Dec 15, 2023 at 2:51
  • $\begingroup$ That's correct. The prime variables are leftover from when I originally wrote to scale everything by 1/S. I'll fix it $\endgroup$
    – qwr
    Dec 15, 2023 at 16:09
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Starting with the special cases out of they way, the way Will Octagon Gibson does, you also get the rest of the solution. Let $x$ be defined by $c=x*a$, so $a+b=x*a+d$ which gives $b=(x-1)*a+d$. The products are $a*b=x*a*d$ which gives $b=x*d$, so $(x-1)*a+d=x*d$ which finally gives $(x-1)*a=(x-1)*d$, so either $x=1$ and therefore $c=a$, or $d=a$

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Let $u = \frac{a+b}2, v = \frac{a-b}2$. Then $2u=a+b$. Also, $a = u+v, b = u-v$, so $ab=u^2-v^2$.

If we define $u'$ and $v'$ similarly in terms of $c,d$, then since $2u'=c+d=a+b=2u$, we can conclude that $u'=u$.

We also have $u'^2+v'^2=u^2+v^2$. Cancelling out $u'^2$ and $u^2$, we get $v'^2=v^2$. This gets us $v'=\pm v$, but we can without loss of generality choose $a \ge b$ and $c \ge d$, which makes $v'$ and $v$ both nonnegative. Hence, $v'=v$.

$c=u'+v'=u+v=a$ and $d=u'-v'=u-vb$, so $(c,d) =(a,b)$.

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Here's an alternate proof which (to me) makes clear why this is true.

Let's simplify and just talk about one pair. Even better, let's assume both members of the pair are equal. Call this value $m$. Then their sum is obviously $2m$ and product $m^2$.

Now, let's begin to separate them from each other. Keeping the symmetry, we pull them apart the same amount $\delta$, giving the pair $(m + \delta, m - \delta)$. Their sum is obviously unchanged, but their product is now $m^2 - \delta^2$.

That's it, QED! The pair's sum determines $m$, the product determines $|\delta|$, and then the pair is fully determined. The only choice is if we use $+\delta$ or $-\delta$, which is simply the difference between $(a,b)$ and $(b,a)$ (if we consider these distinct).

And, since the arithmetic mean is $m$, and the geometric mean $\sqrt{m^2 - \delta^2}$, we incidentally proved the AM-GM inequality as well.

What I like about this proof: It chooses the right perspective, the right coordinate system, and then the proposition becomes self-evident. The problem with $(a,b),(c,d)$ is that it injects a false ordering which obscures things. Instead, we use the (midpoint, gap distance) coordinate system, and the symmetry, the determination, the relationships, are self-evident.

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This is equivalent to a well known and simpler problem:

Find two numbers that have a given sum and product

This was solved by the ancient Babylonians and is, to my knowledge, the first recorded case of a solving a quadratic.

There is one solution (up to order of the two numbers), which implies that in the OP, both pairs are identical (up to order).

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  • $\begingroup$ To be fair not too many records of anything exist before the ancient babylonians. $\endgroup$ Jan 4 at 5:11

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