How many pairs of integers $(x,y)$, with negatives allowed, satisfy $x^2-y^2=45$?

Factorise as $(x-y)(x+y)=45$. Because 45 is odd, both brackets are automatically odd, so every factor pair works and $x,y$ come out whole. Taking the brackets positive in the six possible orders gives (23, 22), (9, 6), (7, 2), (7, −2), (9, −6) and (23, −22). Making both brackets negative flips the sign of $x$ and produces six more: (−23, ±22), (−9, ±6), (−7, ±2). That is 12 pairs in all. The trap 3 counts only the pairs with both $x$ and $y$ positive; 6 allows $y$ to be negative but keeps $x$ positive; and 24 double-counts by swapping $x$ and $y$, which is not allowed since the roles are different.