# 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 (Four Variables)

We consider $f(n)$ and $g(n)$ separately:

• $\boldsymbol{f(n)}$

We restrict $n$ in which $k^2 for some positive integer $k,$ or $$n=(k+1)^2-p\hspace{40mm}(1)$$ for some integer $p$ such that $0\leq p<2k+1.$ By recursion, 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{19mm}(2) \\ \end{align*}

• $\boldsymbol{g(n)}$

If $n$ and $(k+1)^2$ have the same parity, then we get $g(n)=f(n)$ by a similar process from $g\left((k+1)^2\right)=k+1.$ 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.

It follows that $k^2 or $$n=(k+2)^2-2q\hspace{38.25mm}(3)$$ for some integer $q$ such that $0 By recursion, 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{15.5mm}(4) \\ \end{align*}

By $(2)$ and $(4),$ we have $$\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{27mm}(5)$$ From $(1)$ and $(3),$ equating the expressions for $n$ gives $(k+1)^2-p=(k+2)^2-2q.$ Solving for $k$ produces $$k=\frac{2q-p-3}{2}. \hspace{41.25mm}(6)$$ 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{29mm}(7) \end{align*} Since $\gcd(11,10)=1,$ we know that $p-1$ must be divisible by $10,$ and $q$ must be divisible by $11.$

Recall that the restrictions on $(1)$ and $(2)$ are $0\leq p<2k+1$ and $0 respectively. Substituting $(6)$ into either inequality gives $p+1 Combining all these results produces $$0 To minimize $n$ in either $(1)$ or $(3),$ we minimize $k,$ so we minimize $p$ and $q$ in $(8).$ From $(6)$ and $(7),$ we construct the following table: $\[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{k} & \textbf{Satisfies }\boldsymbol{(8)?} \\ [0.5ex] \hline & & & \\ [-2ex] 11 & 11 & 4 & \\ 21 & 22 & 10 & \\ 31 & 33 & 16 & \checkmark \\ \geq41 & \geq44 & \geq22 & \checkmark \\ \end{array}$$ Finally, we have $(p,q,k)=(31,33,16).$ Substituting this result into either $(1)$ or $(3)$ generates $n=\boxed{258}.$

• Remark

We can verify that $$\frac{f(258)}{g(258)}=\frac{1\cdot31+f(258+1\cdot31)}{2\cdot33+g(258+2\cdot33)}=\frac{31+\overbrace{f(289)}^{17}}{66+\underbrace{g(324)}_{18}}=\frac{48}{84}=\frac47.$$

~MRENTHUSIASM

## Solution 3

Since $n$ isn't a perfect square, let $n=m^2+k$ with $0. If $m$ is odd, then $f(n)=g(n)$. If $m$ is even, then \begin{align*} f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\ g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k, \end{align*} from which \begin{align*} 7(3m+2-k)&=4(5m+6-k) \\ m&=3k+10. \end{align*} Since $m$ is even, $k$ is even. Since $k\neq 0$, the smallest $k$ is $2$ which produces the smallest $n$: $$k=2 \implies m=16 \implies n=16^2+2=\boxed{258}.$$ ~Afo

## Solution 4 (Quadratics With Two Variables)

To begin, note that if $n$ is a perfect square, $f(n)=g(n)$, so $f(n)/g(n)=1$, so we must look at values of $n$ that are not perfect squares (what a surprise). First, let the distance between $n$ and the first perfect square greater than or equal to it be $k$, making the values of $f(n+k)$ and $g(n+k)$ integers. Using this notation, we see than $f(n)=k+f(n+k)$, giving us a formula for the numerator of our ratio. However, since the function of $g(n)$ does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of $\sqrt{n+k}$ in $g(n)$ unless $k$ is an even number. However, this is impossible, since if $k$ was an even number, $f(n)=g(n)$, giving a ratio of one. Thus, $k$ must be an odd number.

Thus, since $k$ must be an odd number, regardless of whether $n$ is even or odd, to get an integral value in $g(n)$, we must get to the next perfect square after $n+k$. To make matters easier, let $z^2=n+k$. Thus, in $g(n)$, we want to achieve $(z+1)^2$.

Expanding $(z+1)^2$ and substituting in the fact that $z=\sqrt{n+k}$ yields:

$$(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1$$

Thus, we must add the quantity $k+2z+1$ to $n$ to achieve a integral value in the function $g(n)$. Thus.

$$g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}$$

However, note that the quantity within the square root is just $(z+1)^2$, and so:

$$g(n)=k+3z+2$$

Thus, $$\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}$$

Since we want this quantity to equal $\frac{4}{7}$, we can set the above equation equal to this number and collect all the variables to one side to achieve

$$3k-5z=8$$

Substituting back in that $z=\sqrt{n+k}$, and then separating variables and squaring yields that

$$9k^2-73k+64=25n$$

Now, if we treat $n$ as a constant, we can use the quadratic formula in respect to $k$ to get an equation for $k$ in terms of $n$ (without all the squares). Doing so yields

$$\frac{73\pm\sqrt{3025+900n}}{18}=k$$

Now, since $n$ and $k$ are integers, we want the quantity within the square root to be a perfect square. Note that $55^2=3025$. Thus, assume that the quantity within the root is equal to the perfect square, $m^2$. Thus, after using a difference of squares, we have $$(m-55)(m+55)=900n$$ Since we want $n$ to be an integer, we know that the $LHS$ should be divisible by five, so, let's assume that we should have $m$ divisible by five. If so, the quantity $18k-73$ must be divisible by five, meaning that $k$ leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!).

Thus, we see that to achieve integers $n$ and $k$ that could potentially satisfy the problem statement, we must try the values of $k$ congruent to one modulo five. However, if we recall a statement made earlier in the problem, we see that we can skip all even values of $k$ produced by this modulo argument.

Also, note that $k=1,6$ won't work, as they are too small, and will give an erroneous value for $n$. After trying $k=11,21,31$, we see that $k=31$ will give a value of $m=485$, which yields $n=\boxed{258}$, which, if plugged in to for our equations of $f(n)$ and $g(n)$, will yield the desired ratio, and we're done.

Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with "needs proof". If you can prove it, please add a lemma to the solution doing so :)

-mathislife52

## Solution 5 (Basic Substitutions)

First of all, if $n$ is a perfect square, $f(n)=g(n)=\sqrt{n}$ and their quotient is $1.$ So, for the rest of this solution, assume $n$ is not a perfect square.

Let $a^2$ be the smallest perfect square greater than $n$ and let $b^2$ be the smallest perfect square greater than $n$ with the same parity as $n,$ and note that either $b=a$ or $b=a+1.$ Notice that $(a-1)^2 < n < a^2.$

With a bit of inspection, it becomes clear that $f(n) = a+(a^2-n)$ and $g(n) = b+(b^2-n).$

If $a$ and $n$ have the same parity, we get $a=b$ so $f(n) = g(n)$ and their quotient is $1.$ So, for the rest of this solution, we let $a$ and $n$ have opposite parity. We have two cases to consider.

Case 1: $n$ is odd and $a$ is even

Here, we get $a=2k$ for some positive integer $k.$ Then, $b = 2k+1.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n) = 2k+2m+1$ and $g(n) = 2k+1+2m+1+4k+1 = 6k+2m+3.$

We set $\frac{2k+2m+1}{6k+2m+3}=\frac{4}{7},$ cross multiply, and rearrange to get $6m-10k=5.$ Since $k$ and $m$ are integers, the LHS will always be even and the RHS will always be odd, and thus, this case yields no solutions.

Case 2: $n$ is even and $a$ is odd

Here, we get $a=2k+1$ for some positive ineger $k.$ Then, $b=2k+2.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n)=2k+1+2m+1=2k+2m+2$ and $g(n)=2k+2+2m+1+4k+3 = 6k+2m+6.$

We set $\frac{2k+2m+2}{6k+2m+6} = \frac{4}{7},$ cross multiply, and rearrange to get $5k=3m-5,$ or $k=\frac{3}{5}m-1.$ Since $k$ and $m$ are integers, $m$ must be a multiple of $5.$ Some possible solutions for $(k,m)$ with the least $k$ and $m$ are $(2,5), (5,10), (8,15),$ and $(11,20).$

We wish to minimize $k$ since $a=2k+1.$ One thing to keep in mind is the initial assumption $(a-1)^2 < n < a^2.$

The pair $(2,5)$ gives $a=2(2)+1=5$ and $n=5^2-(2(5)+1)=14.$ But $4^2<14<5^2$ is clearly false, so we discard this case.

The pair $(5,10)$ gives $a=2(5)+1=11$ and $n=11^2-(2(10)+1)=100,$ which is a perfect square and therefore can be discarded.

The pair $(8,15)$ gives $a=2(8)+1=17$ and $n=17^2-(2(15)+1)=258,$ which is between $16^2$ and $17^2$ so it is our smallest solution.

So, $\boxed{258}$ is the correct answer.

~mc21s

## Video Solution

https://youtu.be/tRVe2bKwIY8 ~Mathematical Dexterity