# These roots have roots (but) has no correct solution

In this week's Algebra question, most users made the assumption that the only possible solution occurs when $y = x$, and justified that in a variety of (incorrect) ways. Dobromir bravely spoke out against the crowd, by questioning the validity of certain arguments which didn't make sense to him. Fortune favors the brave, and the solutions were indeed incorrect.

For those who solved it, can you demonstrate that you have actually solved the problem without the fatalistic assumption? Can you properly justify that we must have $y = x$? Or is that even necessary?

Hurry, the community eagerly awaits! Oodles of glory to those with the correct argument.

Please post your solution on the problem, and not here. It is still a live problem. Solutions posted here will be removed.

Update: Peter B, Abhishek S. and C L. have posted correct solutions to this question. However, the arguments only work for a special set of numbers.

How do you solve the general case, namely for $M > N> 0$, find all real solutions to

$\begin{cases} x = \sqrt{y + M} - \sqrt{y + N} \\ y = \sqrt{ x + M} - \sqrt{x + N } \\ \end{cases}$ Note by Calvin Lin
6 years, 11 months ago

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I've added my solution to the solution discussion, which doesn't use any complicated arguments. It also immediately extends to $M>N\geq 0$, and explains how to deal with $N< 0$.

Hint: Consider the algebraic conjugate.

Staff - 6 years, 11 months ago

Existence : * In contrast to my original solution, now consider the *continuous map $f:[0,K] \to [0,K]$ given by $f(x)=\sqrt{x+M}-\sqrt{x+N}$, where $K=\sqrt{M}-\sqrt{N}$. First ensure that it is a valid map (i.e. every point in the domain has an image in the co-domain). This can be seen by observing that the function is decreasing as well as non-negative for all $x\in \mathbb{R}_+$. Thus $g(x)=f(f(x))$ is a valid continuous map from $[0,K] \to [0,K]$. Now the domain of the continuous function $g$ is a convex compact subset of $\mathbb{R}$. Hence we can invoke Brouwer's Fixed Point Theorem to conclude the existence of atleast one fixed point of $f(f(x))$.

- 6 years, 11 months ago

Uniqueness (direct proof) : * We also can have a direct elementary proof of uniqueness of the fixed point once we note that the function $g$ is * concave and $g(0)>0$.
If possible, suppose that the function $g(x)$ has more than one fixed point and $\alpha$ and $\beta, (\beta > \alpha)$ are its first two fixed points on the right side of zero (this tacitly assumes that fixed points of $g$ are at most countable, but this is evident by the fundamental theorem of algebra). Since $g$ is concave, for any $\lambda \in [0,1]$, we have
$g(\lambda \alpha + (1-\lambda)\beta) \geq \lambda g(\alpha) + (1-\lambda) g(\beta) = \lambda \alpha + (1-\lambda) \beta$ i.e. $g(z) > z \hspace{3pt} \forall z\in (\alpha, \beta)$ (since, by the assumption, equality can hold only at the end-points). We also have $g(z) > z \hspace{3pt} \forall z \in [0, \alpha)$ due to continuity of $g$, $g(0)>0$ and $\alpha$ being the first fixed point of $g$.
Now we can choose two points $z_1 \in [0,\alpha)$ and $z_2 \in (\alpha, \beta)$ such that $\alpha$ is a convex combination of $z_1$ and $z_2$, i.e $\exists \mu \in [0,1]$ s.t. $\alpha = \mu z_1+(1-\mu)z_2$. Hence by concavity of $g$, we have $\alpha=g(\alpha) \geq \mu g(z_1) + (1-\mu)g(z_2) > \mu z_1 + (1-\mu)z_2 = \alpha$, a contradiction !

- 6 years, 11 months ago

Uniqueness : It is easy to show (noting the fact that $f$ is decreasing and then using elementary calculus) that $f(f(x))$ is an increasing, concave function. Then this note shows that the fixed point (as noted earlier, we are interested in positive fixed points only) of $f(f(x))$ is indeed unique.

- 6 years, 11 months ago

[Solution removed] Added as comment to Jon's solution.

- 6 years, 11 months ago

[Solution removed] Refer to my comment to Jon's solution for my post.

- 6 years, 11 months ago

Thanks for the solutions! Can you solve the general case?

Note: I believe that the current solutions do not allow us to generalize to all positive values as yet.

Staff - 6 years, 11 months ago

can anyone tell me where to know those rules you use cuz simple looks like i dont know much

- 6 years, 10 months ago

We know $f(f(x)) = x$ does imply $f(x)= f^{-1}(x)$ (for a bijective function, so that it's inverse exists). For polynomials, though which are extremely well-behaved functions, the 2 functions $f(x)$ & $f^{-1}(x)$ are indeed symmetric around the axes, about lines belonging to the family $|y|=|x|$. (Be it along 1st-3rd quadrant about $y=-x$ or along 2nd-4th quadrant about $y=x$). My statement in the answer was "The two given equations are symmetric about $y=x$ which is doubtless. Your counterexample of $y=-x$ just coincides with it's inverse & it is more like an identity function (you're stating one of the mirrors about which we should have been reflecting, so they are excluded, or it can be said $f(x)=-x$ is mirror image of $f^{-1}(x)$ about $y=-x$ though it is vague). $y=\frac{1}{x}$ is not a polynomial. Obviously, all functions won't follow this rule (which is quite applicable for polynomials), because we can easily define functions which don't follow any rule, like Dirichlet's function.

So what we could say For a polynomial function, an inverse function exists which is the mirror image of the polynomial wrt one of the lines $y=\pm x$. So for a function (non-coincident with it's inverse), the only points where it intersects with it's inverse are points on $y=\pm x$.

Please correct me if I am missing something.

- 6 years, 11 months ago

I do not see the relevance of your comment to the problem.

1. The function is $y = \sqrt{ x + 10000} - \sqrt{ x + 100}$ which is not (explicitly) polynomial. In fact, it can be shown that $y = \frac{ x^4 - 2200x^2 + 810000}{4x^2}$ which is clearly not a polynomial function. It has behavior that is similar to $\frac{1}{x}$ when close to the origin.

2. The problem doesn't require an inverse function to exist (as a function).

3. It is possible for 2 (polynomial) functions to not be (completely) symmetric about the line $y = x$, but still have solutions which are not of the form $( \alpha, \alpha)$. For example, solve the system of equations $\begin{cases} 2x = 3y^2 -11y + 12 \\ 2y = 3x^2 - 11y + 12. \end{cases}$ The following are (not necessarily all) solutions: $(1,2), (2,1), (3,3)$.

4. All that we are concerned about, is that there is a particular value of $\alpha$ such that $f( f( \alpha) ) = \alpha$. It does not need to hold for all values of $x$, which is your assumption. For example, in the above problem, I created it by finding a polynomial (through lagrange interpolation) which satisfies $f(1) = 2, f(2) = 1 , f(3) = 3$, which is how I knew that there were those 3 solutions (without knowing if it's the complete set).

Staff - 6 years, 11 months ago