I am a high school student trying my luck and self-studying topics in mathematics I find interesting. While reading through a course in Math for Computer Science, I came across the statement that deciding whether or not an algorithm to solve the SAT problem in polynomial time exists is equivalent to the $P = NP$ problem. I'm curious about why and the textbook doesn't provide an explanation.
I already know what the P = NP problem, what the SAT problem is and what P, NP, etc. are.
I want, if possible, an explanation of why those two problems are equivalent that a high schooler can understand, with effort.
First,it is important to realize that “efficiently solves SAT” must be understood as informal shorthand for “Solves SAT in polynomial time”.
If as you say you know what $P$ and $NP$ are, it should be obvious that $P\subseteq NP$. (If not, leave a comment saying so.)
If we could prove $NP\subseteq P$ also, we would know that $P=NP$. To prove $NP\subseteq P$, we would need to show that for any problem $A$ in $NP$, the problem $A$ was also in $P$; that is, for any problem $A$ in $NP$, that there was a polynomial-time algorithm to solve $A$.
SAT is a member of a class of problems called “NP-complete”. These NP-complete problems are in the following sense the ‘hardest’ problems in NP: if any problem in NP is not in P, then problems in NPC are not in P. conversely, if any NP-complete problem is in P, then all of NP is in P.
The reason this is true is that if $A$ is any problem in $NP$, and if $X$ is NP-complete, then an instance of $A$ can be efficiently rewritten (in polynomial time) to be an instance of $X$ whose answer is the same.
Cook's theorem shows how to do this when $X$ is SAT. You can take an instance $A$ of any problem in $NP$, and quickly (in polynomial time) construct a family of statements which can be satisfied if and only if the answer to $A$ was ‘yes’.
So if you could solve SAT instances in polynomial time, you could solve any problem from $NP$ in polynomial time as well, by transforming it into a SAT instance.
Or put more technically, if SAT were in $P$, then every problem in $NP$ would also be in $P$, because any problem in $NP$ can be reduced to SAT. This would prove $NP \subset P$ and therefore (because $P\subseteq NP$ is obvious) $P=NP$.
Conversely, if we somehow already knew $P=NP$, that would imply $NP\subseteq P$, and since SAT is in $NP$, that would prove that SAT was in $P$.
I hope this is at least a little bit helpful.