Example of situation in which adding inequations introduces an infinity of extraneous solutions and only one true solution.

Question

I've been arguing with someone over the following problem:

Find all $m\in\mathbb{R}$ such that $\exists!(x,y)\in\mathbb{R^2}:\ (y \geq x^2 + m) \wedge (x \geq y^2 + m)$

The solution they presented was this: $$(y \geq x^2 + m) \wedge (x \geq y^2 + m) \implies x + y \geq x^2 + y^2 + 2m \implies \\ 0 \geq x^2 - x + \frac 14 + y^2 - y + \frac 14 + 2m - \frac 12 = \left(x - \frac 12\right)^2 + \left(y - \frac 12\right)^2 + 2m - \frac 12 \implies \\ \frac 12 - 2m \geq \left(x-\frac 12\right)^2 + \left(y-\frac 12\right)^2$$

The RHS is $\geq 0$, so in order for this (and hence the original system too) to have solutions, the LHS must be non-negative.

Now, the person I've argued with claims that for the original system to have a unique solution, the implied inequation must have a unique solution as well, so they found $m=\frac 14$, because otherwise, for $m$ less than that, the inequation has an infinity of solutions and for $m$ greater than that, the inequation has no solutions (which indeed means the system has no solutions). I don't think the claim is true (because implication can generate extraneous solutions and after a bit of debating on it, I got them to direct me to a proof that the system indeed has an infinity of solution for $m < \frac 14$, which, by the way I see this problem, completes the exercise.

Now, what I want is to find an example of a similar problem, where adding inequations leads to an infinity of solutions that don't satisfy the original system except for one.

I've been trying to find one, but I'm not that good at 'problem-developing', to call it that way.

I want to find as many and diverse such examples as possible. In the problem above, for $m < \frac 14$ it is impossible for none of the solutions, perhaps except one, to not satisfy the system. Points of the form $\left(\frac 12 + \epsilon\sqrt{\frac 14 - m}, \frac 12 + \epsilon\sqrt{\frac 14 - m}\right)$ satisfy both the system and the implied inequation, for small enough $\epsilon > 0$. I want to find situations in which the infinite solutions of the implied inequation are not solutions of the system, except for one.

Answer 1

Here's a seemingly trivial setup, but it gets at the crux of what you're trying to express:

• $ x \geq 0 $
• $ -x \geq 0 $

Adding them, we get $ 0 \geq 0 $, which is true for all real numbers $x$.

However, there is clearly only 1 solution to the original conditions, namely $ x = 0 $.

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The idea which you're trying to express is that solutions to the original condition are solutions to the final condition, but not vice versa. So, the correct relevant conclusion is:

If we have a unique solution to the final condition, then this implies that there is at most one solution to the initial condition (but still has to be checked, because in some cases it doesn't. Why? Try to find such an example.).

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Other modifications

2.

• $x \geq 0 $
• $ - x^2 \geq 0 $

• $ x^2 \geq y^2 $
• $ - x^2 \geq y^2$

• $2x^2 \geq y^2 $
• $ - x^2 \geq y^2 $

• $ y \geq x^2 $
• $ -y \geq x^2$

Answer 2

After all, I've managed to come myself with 2 examples starting from the ideas I got from Calvin Lin's answer.

They are two problems like the one in my question where you are supposed to find the values of $m$ for which the system has a unique solution:

$$\left\{\begin{align} &(m+1)x^2 ≤ m^2 \\ &-x^2 ≤ -m \end{align}\right.$$

and the slightly modified version

$$\left\{\begin{align} & 2x ≥ \sqrt 2 \\ & (m+1)x^2 ≤ m^2 \\ & 2x - x^2 ≤ \sqrt 2 - m \end{align}\right.$$

In both these problems, when adding the inequations, you get $mx^2 ≤ m(m-1)$ $[\ast]$ which has a unique solution only for $m=1$. For $m > 1$, there are an infinity of solutions to $x^2 ≤ m-1 \iff |x| ≤ \sqrt{m-1}$, for $0 < m < 1$, we have LHS $≥ 0 >$ RHS, for $m = 0$, any $x\in\mathbb{R}$ is a solution and for $m < 0$, LHS $≤ 0 <$ RHS so again any real $x$ works.

However, for the first exercise: For $m = 1$: the unique solution $x = 0$, does not satisfy the system. For $m > 1$: the system becomes $$\left\{\begin{align} & |x| ≤ \frac{m}{\sqrt{m+1}} \\ & |x| ≥ \sqrt m \end{align}\right. $$, but $\sqrt m ≥ \sqrt{m-1}$ and, as we have seen before, $|x| ≤ \sqrt{m-1}$, so none of the infinite solutions of $[\ast]$ are solutions of the system. For $m = 0$: clearly the system has unique solution $x=0$. Then, for $m < 0$, the second inequality of the system is satisfied. When $-1 < m < 0$, we get infinitely many solutions $|x| ≤ \frac{-m}{\sqrt{m+1}}$ and when $m ≤ -1$, any $x \in \mathbb{R}$ satisfies the system.

So even though $[\ast]$ has unique solution for $m=1$, the system does not have a solution in this case, but instead it has a unique solution for $m=0$. Moreover, none of the infinite solutions of $[\ast]$ for $m > 1$ are solutions for the system.

A similar analysis for the second problems yields alike results, out of them remarkable is the fact that for $m < -\frac 12$, there is an interval of solutions, for $-\frac 12 < m < 0$, there are no solutions, while for $m = -\frac 12$, there is exactly one solution, namely $x=\frac{\sqrt 2}2$.