<s>[INST] Given $a,b,c$ are real numbers such that $a-b=8$ and $ab + c^2 +16 =0$. The numerical value of $a^{2021} + b^{2021} +c^{2021}$ is? I saw this question and was intrigued by it. I wanted to see if what I have so far made sense.
So from the second equation $ab + c^2 +16 =0$ we get $c^2 = -(16+ab)$. Which implies $16+ab\leq0$ or $ab\leq-16$. Then using the first equation $a-b=8$ and solving for $a$ we get $a=8+b$. Then substituting in $a$ in the inequality we get:
$\begin{align}
ab&\leq-16\\
(8+b)b&\leq-16\\
8b+b^2&\leq-16\\
b^2+8b+16&\leq0\\
(b+4)^2&\leq 0\\
\end{align}$
So we must have $b=-4$ for this inequality to be true. Then $a=4$, and $c^2 = 0 \implies c = 0$. Hence
$\begin{align}
a^{2021} + b^{2021} +c^{2021} = 4^{2021} + (-4)^{2021}= 4^{2021} -4^{2021} = 0.
\end{align}$
If anyone sees anything wrong let me know or if there is another solution I'm curious!
 [/INST] Yes, you are correct. The numerical value is $0$, with $a=4,b=-4$ and $c=0$.
How I did it [before opening this post]: First note that for there to be a real solution to the equation $ab+16+c^2=0$, the product $ab$ must satisfy $ab \le -16$. The only $(a,b)$ that satisfies $ab \le -16$ with $a-b=8$ is $a=4,b=-4$, and so $c=0$.
So same idea as yours.
If however the "$16$" were replaced with any number larger, there would be no such $a,b,c$ that simultaneously satisfy both of the resulting constraints, and if the "$16"$ were replaced with anything smaller, there would not only be an infinite number of $a,b,c$ that simultaneously satisfy both of the resulting constraints, but there would also be an infinite number of values that $a^{2021}+b^{2021}+c^{2021}$ could take for such $a,b,c$.
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