AMC10 2019 10A, Problem 3

Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$

Solution: Let $A$ be the age of Ana and $B$ be the age of Bonita. Then, \[A-1 = 5(B-1)\] and \[A = B^2.\] Substituting the second equation into the first gives us \[B^2-1 = 5(B-1).\] By using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$ The answer is $16-4 = 12 \implies \fbox{12}$

Answer: 12