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:
Notice that f(n) is obviously n when n is a perfect square, but for n not being a perfect square consider bounding w2<n<(w+1)2.
In this case, notice that we have that f(n)=1+f(n+1) for non squares and so we can rewrite
f(n)=f(w+1)2+(w+1)2−n=w2+2w+1+w+1−n=w2+3n+2−n
Intuitively, think about each f(n)=1+f(n+1) as a "step". From going from (w+1)2 to n there are (w+1)2−n steps, and if this number of steps is even, g(n) will immediately equal f(n).
So for this ratio to exist, we need that (w+1)2 and n to have different parities, and more specifically for w and n to have the same parities.
After establishing this, let's find g(n).
Let's bound n but a little differently, in particular w2<n<(w+2)2.
Then we'd have that
g(n)=g(w+2)+(w+2)2−n=w2+5w+6−n
Now, after having our initial ratio, namely
g(n)f(n)=74
we rewrite this as
7f(n)=4g(n)→7(w2+3w+2−n)=4(w2+5w+6−n)
3w2+w−10=3n
Notice that the right hand side is divisible by 3, so we need 3w2+w−10≡0mod3→w≡1mod3.
Along with this, since our intial bound was n>w2 notice that 3w2>3n and we need that w−10>0→w>10.
For two solutions that work that have the same parity, we see that regardless of if w is odd or even, that 3w2+w−10 is even and so since that would make n even w has to be even and the smallest solution that works is w=16
which yields n = \boxed