# Difference between revisions of "2021 AIME II Problems/Problem 15"

## Problem

Let $f(n)$ and $g(n)$ be functions satisfying $$f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}$$and $$g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}$$for positive integers $n$. Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$.

## Solution 1

Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2 for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$. Note that this formula also returns the correct value when $n=(k+1)^2$, but not when $n=k^2$. Thus $f(n)=k^2+3k+2-n$ for $k^2.

If $2 \mid (k+1)^2-n$, $g(n)$ returns the same value as $f(n)$. This is because the recursion once again stops at $(k+1)^2$. We seek a case in which $f(n), so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$.

Write $7f(n)=4g(n)$, which simplifies to $3k^2+k-10=3n$. Notice that we want the $LHS$ expression to be divisible by 3; as a result, $k \equiv 1 \pmod{3}$. We also want n to be strictly greater than $k^2$, so $k-10>0, k>10$. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is $k=16$, giving $n=258$.

Indeed - if we check our answer, it works. Therefore, the answer is $\boxed{258}$.

-Ross Gao

## Solution 2 (More Variables)

We restrict $n$ in which $k^2 for some positive integer $k,$ or $$n=(k+1)^2-p\hspace{15mm}(1)$$ for some nonnegative integer $p.$ By observations, we get \begin{align*} f\left((k+1)^2\right)&=k+1, \\ f\left((k+1)^2-1\right)&=k+2, \\ f\left((k+1)^2-2\right)&=k+3, \\ &\cdots \\ f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{15mm}(2) \\ \end{align*} If $n$ and $(k+1)^2$ have the same parity, then starting with $g\left((k+1)^2\right)=k+1,$ we get $g(n)=f(n)$ by a similar process. This contradicts the precondition $\frac{f(n)}{g(n)} = \frac{4}{7}.$ Therefore, $n$ and $(k+1)^2$ must have different parities, from which $n$ and $(k+2)^2$ must have the same parity.

Along with the earlier restriction, note that $k^2 or $$n=(k+2)^2-2q\hspace{15mm}(3)$$ for some positive integer $q.$ By observations, we get \begin{align*} g\left((k+2)^2\right)&=k+2, \\ g\left((k+2)^2-2\right)&=k+4, \\ g\left((k+2)^2-4\right)&=k+6, \\ &\cdots \\ g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15mm}(4) \\ \end{align*} By $(2)$ and $(4),$ we have $$\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{15mm}(5)$$ From $(1)$ and $(3),$ equating the expressions for $n$ gives $(k+1)^2-p=(k+2)^2-2q.$ Solving for $k$ produces $$k=\frac{2q-p-3}{2}. \hspace{15mm}(6)$$ We substitute $(6)$ into $(5),$ then simplify, cross-multiply, and rearrange: \begin{align*} \frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\ \frac{p+2q-1}{-p+6q+1}&=\frac{4}{7} \\ 7p+14q-7&=-4p+24q+4 \\ 11p-11&=10q \\ 11(p-1)&=10q. \hspace{15mm}(7) \end{align*} Since $\gcd(11,10)=1,$ we know that $p-1$ must be divisible by $10,$ and $q$ must be divisible by $11.$ Furthermore, rewriting the restriction $k^2 in terms of $p$ and $q$ gives \begin{align*} k^2&<\underbrace{(k+1)^2-p}_{\text{by }(1)} \\ p&<2k+1 \\ p&<2\biggl(\phantom{ }\underbrace{\frac{2q-p-3}{2}}_{\text{by }(6)}\phantom{ }\biggr)+1 \\ p&<(2q-p-3)+1 \\ p&

SOLUTION IN PROGRESS. NO EDIT PLEASE--A MILLION THANKS.

Remark

We can verify that $$\frac{f(258)}{g(258)}=\frac{31+\overbrace{f(289)}^{17}}{2\cdot33+\underbrace{f(324)}_{18}}=\frac{48}{84}=\frac47.$$

~MRENTHUSIASM