Confusion regarding Quadratic equations and RHS = 0

Question

Recently, I am becoming confused with how it is said that in a quadratic equation you MUST make the RHS $ = 0$. But I am stumbling across many equations where it is calculated (The following is for integer solutions) as e.g. $$2^{2x}-3^{2y} = 55$$ $$(2^x-3^y)(2^x+3^y)=55$$ so $$2^x-3^y = 5$$ or $$2^x-3^y = 11$$ Why is this possible? Isn't this wrong? How could you have a theorem that says $a\cdot b = c\cdot d$ so $a=c$ or $a=d$? What I am understanding is that you can also do this: $$(x+4)(x-2) = 10$$ so $$(x+4)(x-2) = 5\cdot 2$$ so $(x+4) = 5$ OR $(x-2)=2$, But this is not correct! So how is it allowed that we can do this for the first equation that I mentioned? And never mind the fact that if you took the 55 over in the first equation you would not be able to simplify it, I simply want an explanation to the arithmetic and logic taking place. Any help and explanations will be greatly appreciated. Thanks.

Answer 1

(I originally posted this as a comment because MSE was, mysteriously, not allowing me to post answers. That has been resolved so I am posting it here now and have deleted the comments.) The solution you present as one you are "stumbling across" is incorrect, for exactly the reasons you identify -- unless there is some context that specifies, for example, that the values of $x$ and $y$ must be positive integers.

In that case, we can use the fact that since $2^x - 3^y$ and $2^x + 3^y$ are both integers, and 5 is a prime, since 5 divides the product of the two integers it must divide (at least) one of them; likewise since 11 is prime and divides the product, it must divide (at least) one of them; and since 55 has no other prime factors, neither $2^x-3^y$ nor $2^x+3^y$ contains any extra prime factors. Also, we observe that neither of them can be equal to 1. So the only way to have two positive integers multiply to be 55 is for one of them to be 5 and the other one to be 11.

In other words: In a number theory context, if we are limiting ourselves to integer solutions to problems, then there are additional heuristics that apply.