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I am trying to find $x+y$ given that

$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$

It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side

$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$

and see if a term $x+y$ comes out, but it looks hopeless at this point.

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    $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. Please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – 6005 May 14 '13 at 13:03
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    $\begingroup$ This is a very nice problem, and one of the best one-line algebra puzzles I have seen on this site. Thank you for posting it. The down-votes are probably the result of people not understanding the problem. (Hints for the automatic negative voters: $x+y$ is supposed to be constant on a nonlinear plane curve? Really? How is that supposed to happen in the world of trivial homework algebra? Seeing the answers, what possible "work" could have been displayed that is not equivalent to having a solution? Are clever, interesting problems like this the kind of thing you want to discourage on MSE?) $\endgroup$ – zyx May 15 '13 at 15:52
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    $\begingroup$ I edited and added an answer to propose a reconsideration of the downvotes. Right now this is at a very curious 10 votes up, 10 votes down. $\endgroup$ – zyx May 23 '13 at 0:00
  • $\begingroup$ Interestingly, this is an amazingly intriguing pre-calculus question and fairly difficult to the naked eye. $\endgroup$ – Mohammad Zuhair Khan Sep 8 '18 at 4:51
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I want to explain why this is actually a remarkable problem.

The question is to show that an algebraic curve, of degree that could potentially be as high as 8 (it is a product of two terms with square roots) defines a straight line -- at least for its real solutions.

The usual ways in which such a thing happens, and are represented fairly often in competition problems and book exercises, are:

  • the equation represents the equality condition in an inequality of real numbers (for example, arithmetic mean equals geometric mean for some suitably constructed set of variables), or

  • the equation defining the curve factorizes, with some of the factors having no real solutions (for example, our degree 8 curve might be $(x+y)^4$ multiplied by a degree $4$ polynomial that is positive for all real coordinates $(x,y)$).

Neither of these is the case here. Algebraic calculation, separating $x$ and $y$ to different sides of the equation, shows that we have $f(x) = f(-y)$ for the increasing function $f(x) = x + \sqrt{x^2+3}$, so that for real solutions $x = -y$. This is a nice argument, and it generalizes to $F(x) = x + \sqrt{x^2 + a^2}$ and the equation $F(x)F(y)=a^2$, but does it fit into some algebraic framework? The issue is what happens with the complex solutions, or when the equation is handled purely algebraically.

Introducing variables $X$ and $Y$ with $X^2 - x^2=3$ and $Y^2 - y^2=3$, there is a system of 3 quadratic equations in 4 unknowns. The surprise here is that purely algebraic calculations on the system in the polynomial ring $\mathbb{Z}[x,y,X,Y]$, lead to $3(x+y)=0$. This illustrates some of the complexities around Bezout's theorem; for affine and singular varieties you cannot naively determine the degree of an intersection by degree counting alone. The algebra is not hard in this case, but it is well worth going through the geometric, projective and scheme-theoretic descriptions of what is happening in this deceptively simple exercise.

I will edit the question to see if any of the 10 downvoters would like to change their minds in light of this information (assuming a non OP edit allows that).

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  • $\begingroup$ A nice answer, and a well deserved +1 for that. I'm not sure about the wisdom of including the last paragraph given that this question has spawned a meta thread of its own. $\endgroup$ – Jyrki Lahtonen May 27 '13 at 5:40
  • $\begingroup$ @JyrkiLahtonen, thanks. I posted this because it spawned the meta thread, to show that the conviction of (in this case) 10 downvoters, about half of whom are also close voters, can mean exactly nothing about actual quality of questions. $\endgroup$ – zyx May 27 '13 at 6:52
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    $\begingroup$ I realize that. I agree that throwing a PSQ comment template at this question was not best (see my reply to Robjohn's comment in meta). But I don't think handing the answer on a platter is a good practice here either - even though I refrained from downvoting the posted full solutions. $\endgroup$ – Jyrki Lahtonen May 27 '13 at 6:56
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$$ \begin{align} x+\sqrt{x^2+3} &=\frac3{y+\sqrt{y^2+3}}\\ &=\frac3{y+\sqrt{y^2+3}}\frac{y-\sqrt{y^2+3}}{y-\sqrt{y^2+3}}\\ &=\frac3{-3}\left(y-\sqrt{y^2+3}\right)\\ &=-y+\sqrt{y^2+3}\tag{1} \end{align} $$ Similarly $$ y+\sqrt{y^2+3}=-x+\sqrt{x^2+3}\tag{2} $$ Add $(1)$ and $(2)$ and cancel the radicals.

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    $\begingroup$ Would the downvoter care to comment? If this is a PSQ matter (which I would doubt since my answer is less complete than the others which were not downvoted), please read my comment here. $\endgroup$ – robjohn May 15 '13 at 7:35
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    $\begingroup$ Piling down/close votes on clever problems and illuminating answers in the name of the great warrior god "Psq" does not strike me as a good development strategy for MSE. I have added a comment under the question about this, and agree with your remark in the meta thread. $\endgroup$ – zyx May 15 '13 at 16:07
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    $\begingroup$ This is a very slick solution, +1 $\endgroup$ – Matthew Towers May 27 '13 at 10:41
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Hint: Calculate $(-t+\sqrt{t^2+3})(t+\sqrt{t^2+3})$.

Details: The product in the hint is $3$. So for every $y$, by putting $x=-y$ we get a solution of the equation $$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3}=3)=3\tag{$1$}$$ with $x+y=0$. We now show these are the only solutions,

One of the terms in $(1)$ is $\le \sqrt{3}$, say the first one. Then the second one is $\ge \sqrt{3}$. But for positive $t$, the function $t+\sqrt{t^2+3}$ is increasing, so there is a unique value of $y$ corresponding to $x$.

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    $\begingroup$ concise, but arcane (+1) $\endgroup$ – robjohn May 14 '13 at 13:26
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I want to add another perspective to this problem. Not from the point of view of algebraic varieties, see zyx's answer for that, but from the point of view of teaching a problem solving strategy to beginners. The solution (as seen from several posts) depends on a useful trick of clearing the square roots from the denominator, but IMVHO there are important steps before we reach that point.

Let's take a look at the problem. We are given a single equation tying the values of two variables $x$ and $y$ together. Fine - business as usual. The equation looks a bit scary because of the square roots and all that, but presumably such an equation defines a curve of some kind. But we are asked about the value of another function, a much simpler one, the sum of the two coordinates. Wait one red minute!? What is this? It sounds as if a single number is expected? May be a range of values? No, that wouldn't quite match the chosen wording. As it stands the question is a bit strange, for a priori there is little reason to suspect that we could say much at all about the value of $x+y$. But let's keep an open mind!

None of the elementary things seem to say much about the equation we were given. How could we even get started? It's time for my battle cry: I couldn't make sense out of it, so I calculated some examples! Let's plug in some values for $x$, solve for $y$, and see how the land lies. A physicist would call this step "experimental" or "making observations" - the technique is somewhat underrated in math education (though professionals do it all the time). Let's try something simple. What if $x=0$? Then $x+\sqrt{x^2+3}=\sqrt3$, and we are to solve $y+\sqrt{y^2+3}=3/\sqrt3=\sqrt3.$ Hmm? A bit ugly. Let's try $x=1$. Then we get $x+\sqrt{x^2+3}=1+\sqrt{4}=3$, and we are to solve $y+\sqrt{y^2+3}=1$. This time the value on the r.h.s. was a clean integer, which is an improvement I guess, but no flashes of insight yet. Any other values of $x$ that would make $x+\sqrt{x^2+3}$ simple? $x=-1$ comes to mind given that again $\sqrt{x^2+3}$ will be equal to two. So this time $x+\sqrt{x^2+3}=-1+2=1$, and we need to solve $y+\sqrt{y^2+3}=1$.

Ok, to get any example points $(x,y)$ we need to solve equations of the form $$ y+\sqrt{y^2+3}=a $$ for some $a$. How do we do that? Well, this is something where we can apply things about equations that we have learned earlier. Keep the square root on one side, move the rest to the other, and square both sides. We need to keep track of the signs, but that's routine. So we get as a consequence $$ y^2+3=\left(\sqrt{y^2+3}\right)^2=(a-y)^2=a^2-2ay+y^2. $$ Let's see, the $y^2$ terms cancel, so we get $2ay=a^2-3$ and $y=(a^2-3)/(2a)$. When $x=1$, we had $a=1$, so $y=(1^2-3)/(2\cdot1)=-1$. Similarly, when $x=-1$ we have $a=3$ and $y=(3^2-3)/6=1$. And when $x=0$, we have $a=\sqrt3$ and $y=(3-3)/2a=0$.

So we at long last have observed some data points: $(x,y)$ can be one of $(1,-1)$, $(0,0)$ or $(-1,1)$. There are countless other points, and you are advised to produce a few others, if this is not enough. What was the question again? The value of $x+y$? Hmm, at all these "observed" points the coordinates sum to zero. Coincidence? Possibly, we are math folks, so suspicion is second nature. But let's try this further (for lack of better ideas, really). If we really have $x+y=0$ always, then we should have $y=-x$. Does that fit? Let's check $$ (x+\sqrt{x^2+3})(-x+\sqrt{(-x)^2+3})=(x+\sqrt{x^2+3})(-x+\sqrt{x^2+3})?? $$ Ahh, it's the good ol' sum of two things $\times$ their difference: $(a+b)(a-b)=a^2-b^2$ with $a=\sqrt{x^2+3}$ and $b=x$. We have $a^2-b^2=(x^2+3)-x^2=3$! Success! Success? We just showed that points on the line $x+y=0$ do satisfy the given equation, but the direction of the implication is wrong. We wanted the converse. Leaving that to you! Insert a devilish grin here, but also the hint that now you may peek at the other solutions. Also our calculations above showed that to each $x$ there is a single $a$ (show that you never need to divide by zero!) and to each $a$ there will be a single $y$. We already know which $y$ will work at least, so it looks like there are no others???


This is by no means the only approach, but it may take you the distance, when you are in unfamiliar territory. The successful answerers apparently drew from their personal databanks and tuned up pattern recognition heuristics. Of course, building up the said databanks takes time, and you need to work your way through scores of examples to develop and improve those pattern recognition skills.

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    $\begingroup$ Of course, after you have done all this, and built up a solution, it is time to follow Gauss' advice and remove all the scaffolding to leave a beautiful sculpture or a piece of architecture standing on its own. All those meandering paths of thought removed from the scene leaving only esthetically pleasing things for the onlookers. $\endgroup$ – Jyrki Lahtonen May 27 '13 at 6:49
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$$ p = x + \sqrt{x^2+3} \Rightarrow \sqrt{x^2+3} = p-x \\ p^2 = x^2 + x^2+3 + 2x \sqrt{x^2+3} = 2x^2+3 + 2x(p-x) = 2xp+3 \\ x = \frac {p^2-3}{2p} $$ Analogously, if $y + \sqrt{y^2+3} = q$, then $$ y = \frac {q^2-3}{2q} $$ Initial expressions becomes $pq = 3$. Now, if calculate $x+y$ you'll get $$ x+y = \frac {p^2-3}{2p} + \frac {q^2-3}{2q} = \frac 12 \left (\frac {p^2q-3q+q^2p-3p}{pq} \right ) = \frac {p+q}2 \left( pq-3\right) = 0 $$

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Alt. hint:   following up on OP's $^{\tiny\text{(?)}}\,$ approach, with the missing $\,=3\,$ appended...

I tried to expand the left hand side: $$\;xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)} = 3 \tag{1}$$

Before that, multiplying the original equation by $\,\left(x\color{red}{-}\sqrt{x^2+3}\right)\left(y\color{red}{-}\sqrt{y^2+3}\right)\,$ gives:

$$ (-3)\,(-3) = 3 (x-\sqrt{x^2+3})(y-\sqrt{y^2+3}) \;\;\iff\;\; (x-\sqrt{x^2+3})(y-\sqrt{y^2+3}) = 3 \tag{2} $$

Expanding $(2)$ and subtracting from $(1)$ then gives:

$$\require{cancel} \begin{align} y\sqrt{x^2+3} + x\sqrt{y^2+3} = 0 \;\;&\iff\;\; y\sqrt{x^2+3} = -x\sqrt{y^2+3} \\ &\implies\;\;y^2(\cancel{x^2}+3)=x^2(\cancel{y^2}+3) \\ &\iff\;\; x^2 = y^2 \\ &\iff\;\; (x-y)(x+y) = 0 \end{align} $$

The case $\,x-y = 0\,$ has the only solution $\,x=y=0\,$, which leaves $\,x+y=0\,$ either way.

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