We can change the bounded condition to \(|a - b| \leq 1\) for all \(a, b \in A\). It is clear that \(|a - b| \leq 1\) implies boundness, and, if \(a - b > 1\) for some \(a, b \in A\), then \((a - b)^{2^n} \in A\), which implies \(A\) is unbounded. (I am using the fact that, if \(A\) is non-empty, then \(0 \in A\).)

Solution during the test. Suppose that exist a set $A$ with $x \in A$ and $x < \frac{-1}{4}$. Define the sequence $b_n$ as $b_1=0$ and $b_n = (b_{n-1}-x)^2$. Clearly by induction we have that $0= b_1 = (x-x)^2 \in A$ and $b_n \in A$. Now there exists $\epsilon $ such that $b_{n+1} = b_n + \epsilon$. Let $x=-y$ and $y>\frac{1}{4}$. We have: $$ b_{n+1} = (b_n+y)^2 = b_n + \epsilon \Longrightarrow b_n^2+(2y-1)b_n+y^2=\epsilon $$Looking to this quadratic, we have $\Delta = (2y-1)^2-4y^2= 1-4y <0$, hence this quadratic has a positive minimum. Hence for this minimum $\epsilon_0$ we have $b_n \geq \epsilon_0+b_{n-1}$ for all $n$ $ \Longrightarrow b_n$ is unbounded. Therefore $A$ also is unbounded, which is a contradiction. A example for $\frac{-1}{4}$ is the interval $(\frac{-1}{4}, \frac{1}{4})$, and is easy to check that this works. $\blacksquare$

Note that $[-1/4,1/4]$ is framed. Suppose $x<\frac{-1}{4}$ works. Let $S$ be the supremum of A and $(x_i)$ be a sequence of elements of $A$ that converges to $S$. Then, $$(x_i-x)^2\le S$$Taking the limit as $i\to \infty$, $$(S+\frac{1}{4})^2<(S-x)^2\le S$$$$(S-1/4)^2<0,$$impossible. $\blacksquare$

ZeusDM wrote: A set \(A\) of real numbers is framed when it is bounded and, for all \(a, b \in A\), not necessarily distinct, \((a-b)^{2} \in A\). What is the smallest real number that belongs to any framed set? There is an translation error in your post; the "any" in your message means "qualquer" and not "algum". So the way you have written is equivalent to "What is the smallest real number that belongs to all framed set?". The original problem says: "What is the smallest real number that belongs to some framed set?".

Note that $[-\frac{1}{4}; \frac{1}{4}]$ is framed. Now we claim that $-\frac{1}{4}$ is in fact the smallest real that is an element of a framed set. Suppose there exists a framed set $[-(\frac{1}{4}+\epsilon);b]$, $\epsilon$ a non-negative real. Note that in fact $(-(\frac{1}{4}+\epsilon)-b)^2=(\frac{1}{4}+\epsilon+b)^2 \geq (\frac{1}{4}+b)^2 \geq b$, where equality only holds for $b=\frac{1}{4}$ and $\epsilon=0$. But we also want to have a bounded set and therefore we must have $(-(\frac{1}{4}+\epsilon)-b)^2 \leq b$. Thus $b=\frac{1}{4}$ and $\epsilon=0$ and hence $-\frac{1}{4}$ is indeed our smallest real being an element of a framed set.