Problem:
Let f(n) and g(n) be functions satisfying
f(n)={n1+f(n+1) if n is an integer otherwise
and
g(n)={n2+g(n+2) if n is an integer otherwise
for positive integers n. Find the least positive integer n such that g(n)f(n)=74.
Solution:
Answer (258):
For a given n, let k be the least positive integer such that k2≥n. Then
f(n)=1+f(n+1)=2+f(n+2)=⋯=(k2−n)+f(k2)=(k2−n)+k.
In other words, to compute f(n), find the least perfect square greater than or equal to n, and then f(n) is the sum of the square root of that perfect square and the distance from n to the perfect square. This also holds for g(n), but instead of finding the least perfect square greater than or equal to n, find the least perfect square greater than or equal to n that has the same parity as n.\
If n and the next greater perfect square k2 have the same parity, then f(n)=g(n), and the required condition cannot hold. Hence n and k must have different parities, so f(n)=k2−n+k and g(n)=(k+1)2−n+k+1. If g(n)f(n)=74, then
(k+1)2−n+k+1k2−n+k=74
which simplifies to 3n=3k2−5k−8. Therefore k≡2(mod3). Also, because 3k2−5k is always even, it follows that n is even and k must be odd.\
Because k2 must be the least perfect square with k2≥n, it follows that n≥(k−1)2+1. Substituting for n in terms of k gives
k2−35k−38≥k2−2k+2
which is equivalent to k≥14. The least k≥14 that is odd and satisfies k≡2(mod3) is k=17, which gives n=172−35⋅17−38=258.