AMC10 2023 B

Let's use 10th grade math to solve this. After all, it is called the AMC 10 for a reason! We have \[m^2 + mn + n^2 = m^2n^2.\] We subtract mn on both sides to get m2+n2=m2n2−mn. Fun Fact! You can write m2+n2 as (m+n)2−2mn! Let's use this! We convert the Left Hand Side into (m+n)2−2mn to get (m+n)2−2mn=m2n2−mn. Adding by 2mn gives us (m+n)2=m2n2+mn. We aren't done yet though! m2n2+mn can be simplified into mn(mn+1), giving us (m+n)2=mn(mn+1). Okay so now we're done, but, Pinotation, this doesn't do anything! Well, now we can use the Zero Product Property! How? I'll show you! We subtract mn(mn+1) on both sides of the equation to get (m+n)2−mn(mn+1)=0. Now we do a bit of casework. Notice how (m+n)2−mn(mn+1)=0 is just (m+n)(m+n)−mn(mn+1)=0. So, either m+n=0 and mn=0, or m+n=0 and mn+1=0. Let's look at it through both cases. Case 1: m+n=0 and mn=0. If m+n=0 and mn=0, then that must mean that either m=0 or n=0, and if we substitute either m=0 or n=0 in, we still get either m=0 or n=0, so therefore we have 1 ordered pair, (0,0). Case 2: m+n=0 and mn+1=0. mn+1=0 means that mn=−1. So, for this to be possible, either m=−1 and n=1, or m=1 and n=−1. Let's check for contradictions quickly. We see that m+n=0, and −1+1=0 and 1−1=0, so we know the ordered pairs (−1,1) and (1,−1) both work. We have a total of $\boxed{\textbf{(C) 3}}.$ ordered pairs. (−1,1), (0,0), and (1,−1).