Please explain the solution to the following Scheme programming problem:

2. Write a function square-factor that takes one argument n. This function determines if there is any square integer that is a factor of n. For example, (square-factor 3) is #f; (square-factor 4) is #t since 22 divides evenly into 4. Also, (square-factor 27) is #t since 32 divides evenly into 27, but (square-factor 10) is #f. s #t if there is some integer k where k2 evenly divides into n (square-factor n) = { #fotherwise Solution: (define (square-factor n) (define (square x) (* x x)) (define (helper nk) (cond ((= k 1) #f) ((= 0 (remainder n (square k))) #t) (else (helper n (- k 1))))) (helper n (- n 1))) 2. Write a function square-factor that takes one argument n. This function determines if there is any square integer that is a factor of n. For example, (square-factor 3) is #f; (square-factor 4) is #t since 22 divides evenly into 4. Also, (square-factor 27) is #t since 32 divides evenly into 27, but (square-factor 10) is #f. s #t if there is some integer k where k2 evenly divides into n (square-factor n) = { #fotherwise Solution: (define (square-factor n) (define (square x) (* x x)) (define (helper nk) (cond ((= k 1) #f) ((= 0 (remainder n (square k))) #t) (else (helper n (- k 1))))) (helper n (- n 1)))