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

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m (Solution 5 (Basic Substitutions))
 
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\end{align*}</cmath></li><p>
 
\end{align*}</cmath></li><p>
 
   <li><math>\boldsymbol{g(n)}</math><p>
 
   <li><math>\boldsymbol{g(n)}</math><p>
If <math>n</math> and <math>(k+1)^2</math> have the same parity, then starting from <math>g\left((k+1)^2\right)=k+1,</math> we get <math>g(n)=f(n)</math> by a similar process. This contradicts the precondition <math>\frac{f(n)}{g(n)} = \frac{4}{7}.</math> Therefore, <math>n</math> and <math>(k+1)^2</math> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. <p>
+
If <math>n</math> and <math>(k+1)^2</math> have the same parity, then we get <math>g(n)=f(n)</math> by a similar process from <math>g\left((k+1)^2\right)=k+1.</math> This contradicts the precondition <math>\frac{f(n)}{g(n)} = \frac{4}{7}.</math> Therefore, <math>n</math> and <math>(k+1)^2</math> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. <p>
 
It follows that <math>k^2<n<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{38.25mm}(3)</cmath> for some integer <math>q</math> such that <math>0<q<2k+2.</math> By recursion, we get
 
It follows that <math>k^2<n<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{38.25mm}(3)</cmath> for some integer <math>q</math> such that <math>0<q<2k+2.</math> By recursion, we get
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 68: Line 68:
 
Finally, we have <math>(p,q,k)=(31,33,16).</math> Substituting this result into either <math>(1)</math> or <math>(3)</math> generates <math>n=\boxed{258}.</math></li><p>
 
Finally, we have <math>(p,q,k)=(31,33,16).</math> Substituting this result into either <math>(1)</math> or <math>(3)</math> generates <math>n=\boxed{258}.</math></li><p>
 
   <li><b>Remark</b><p>
 
   <li><b>Remark</b><p>
We can verify that <cmath>\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{f(324)}_{18}}=\frac{48}{84}=\frac47.</cmath></li><p>
+
We can verify that <cmath>\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.</cmath></li><p>
 
</ul>
 
</ul>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
==Solution 3==
 +
Since <math>n</math> isn't a perfect square, let <math>n=m^2+k</math> with <math>0<k<2m+1</math>. If <math>m</math> is odd, then <math>f(n)=g(n)</math>. If <math>m</math> is even, then
 +
<cmath>\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*}</cmath>
 +
from which
 +
<cmath>\begin{align*}
 +
7(3m+2-k)&=4(5m+6-k) \\
 +
m&=3k+10.
 +
\end{align*}</cmath>
 +
Since <math>m</math> is even, <math>k</math> is even. Since <math>k\neq 0</math>, the smallest <math>k</math> is <math>2</math> which produces the smallest <math>n</math>: <cmath>k=2 \implies m=16 \implies n=16^2+2=\boxed{258}.</cmath>
 +
~Afo
 +
 +
==Solution 4 (Quadratics With Two Variables)==
 +
To begin, note that if <math>n</math> is a perfect square, <math>f(n)=g(n)</math>, so <math>f(n)/g(n)=1</math>, so we must look at values of <math>n</math> that are not perfect squares (what a surprise). First, let the distance between <math>n</math> and the first perfect square greater than or equal to it be <math>k</math>, making the values of <math>f(n+k)</math> and <math>g(n+k)</math> integers. Using this notation, we see than <math>f(n)=k+f(n+k)</math>, giving us a formula for the numerator of our ratio. However, since the function of <math>g(n)</math> 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 <math>\sqrt{n+k}</math> in <math>g(n)</math> unless <math>k</math> is an even number. However, this is impossible, since if <math>k</math> was an even number, <math>f(n)=g(n)</math>, giving a ratio of one. Thus, <math>k</math> must be an odd number.
 +
 +
Thus, since <math>k</math> must be an odd number, regardless of whether <math>n</math> is even or odd, to get an integral value in <math>g(n)</math>, we must get to the next perfect square after <math>n+k</math>. To make matters easier, let <math>z^2=n+k</math>. Thus, in <math>g(n)</math>, we want to achieve <math>(z+1)^2</math>.
 +
 +
Expanding <math>(z+1)^2</math> and substituting in the fact that <math>z=\sqrt{n+k}</math> yields:
 +
 +
<cmath>(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1</cmath>
 +
 +
Thus, we must add the quantity <math>k+2z+1</math> to <math>n</math> to achieve a integral value in the function <math>g(n)</math>. Thus.
 +
 +
<cmath>g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}</cmath>
 +
 +
However, note that the quantity within the square root is just <math>(z+1)^2</math>, and so:
 +
 +
<cmath>g(n)=k+3z+2</cmath>
 +
 +
Thus,
 +
<cmath>\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}</cmath>
 +
 +
Since we want this quantity to equal <math>\frac{4}{7}</math>, we can set the above equation equal to this number and collect all the variables to one side to achieve
 +
 +
<cmath>3k-5z=8</cmath>
 +
 +
Substituting back in that <math>z=\sqrt{n+k}</math>, and then separating variables and squaring yields that
 +
 +
<cmath>9k^2-73k+64=25n</cmath>
 +
 +
Now, if we treat <math>n</math> as a constant, we can use the quadratic formula in respect to <math>k</math> to get an equation for <math>k</math> in terms of <math>n</math> (without all the squares). Doing so yields
 +
 +
<cmath>\frac{73\pm\sqrt{3025+900n}}{18}=k</cmath>
 +
 +
Now, since <math>n</math> and <math>k</math> are integers, we want the quantity within the square root to be a perfect square. Note that <math>55^2=3025</math>. Thus, assume that the quantity within the root is equal to the perfect square, <math>m^2</math>. Thus, after using a difference of squares, we have
 +
<cmath>(m-55)(m+55)=900n</cmath>
 +
Since we want <math>n</math> to be an integer, we know that the <math>LHS</math> should be divisible by five, so, let's assume that we should have <math>m</math> divisible by five. If so, the quantity <math>18k-73</math> must be divisible by five, meaning that <math>k</math> 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 <math>n</math> and <math>k</math> that could potentially satisfy the problem statement, we must try the values of <math>k</math> 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 <math>k</math> produced by this modulo argument.
 +
 +
Also, note that <math>k=1,6</math> won't work, as they are too small, and will give an erroneous value for <math>n</math>. After trying <math>k=11,21,31</math>, we see that <math>k=31</math> will give a value of <math>m=485</math>, which yields <math>n=\boxed{258}</math>, which, if plugged in to for our equations of <math>f(n)</math> and <math>g(n)</math>, 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 <math>n</math> is a perfect square, <math>f(n)=g(n)=\sqrt{n}</math> and their quotient is <math>1.</math> So, for the rest of this solution, assume <math>n</math> is not a perfect square.
 +
 +
Let <math>a^2</math> be the smallest perfect square greater than <math>n</math> and let <math>b^2</math> be the smallest perfect square greater than <math>n</math> with the same parity as <math>n,</math> and note that either <math>b=a</math> or <math>b=a+1.</math> Notice that <math>(a-1)^2 < n < a^2.</math>
 +
 +
With a bit of inspection, it becomes clear that <math>f(n) = a+(a^2-n)</math> and <math>g(n) = b+(b^2-n).</math>
 +
 +
If <math>a</math> and <math>n</math> have the same parity, we get <math>a=b</math> so <math>f(n) = g(n)</math> and their quotient is <math>1.</math> So, for the rest of this solution, we let <math>a</math> and <math>n</math> have opposite parity. We have two cases to consider.
 +
 +
Case 1: <math>n</math> is odd and <math>a</math> is even
 +
 +
Here, we get <math>a=2k</math> for some positive integer <math>k.</math> Then, <math>b = 2k+1.</math> We let <math>n = a^2-(2m+1)</math> for some positive integer <math>m</math> so <math>f(n) = 2k+2m+1</math> and <math>g(n) = 2k+1+2m+1+4k+1 = 6k+2m+3.</math>
 +
 +
We set <math>\frac{2k+2m+1}{6k+2m+3}=\frac{4}{7},</math> cross multiply, and rearrange to get <math>6m-10k=5.</math> Since <math>k</math> and <math>m</math> are integers, the LHS will always be even and the RHS will always be odd, and thus, this case yields no solutions.
 +
 +
Case 2: <math>n</math> is even and <math>a</math> is odd
 +
 +
Here, we get <math>a=2k+1</math> for some positive ineger <math>k.</math> Then, <math>b=2k+2.</math> We let <math>n = a^2-(2m+1)</math> for some positive integer <math>m</math> so <math>f(n)=2k+1+2m+1=2k+2m+2</math> and <math>g(n)=2k+2+2m+1+4k+3 = 6k+2m+6.</math>
 +
 +
We set <math>\frac{2k+2m+2}{6k+2m+6} = \frac{4}{7},</math> cross multiply, and rearrange to get <math>5k=3m-5,</math> or <math>k=\frac{3}{5}m-1.</math> Since <math>k</math> and <math>m</math> are integers, <math>m</math> must be a multiple of <math>5.</math> Some possible solutions for <math>(k,m)</math> with the least <math>k</math> and <math>m</math> are <math>(2,5), (5,10), (8,15),</math> and <math>(11,20).</math>
 +
 +
We wish to minimize <math>k</math> since <math>a=2k+1.</math> One thing to keep in mind is the initial assumption <math>(a-1)^2 < n < a^2.</math>
 +
 +
The pair <math>(2,5)</math> gives <math>a=2(2)+1=5</math> and <math>n=5^2-(2(5)+1)=14.</math> But <math>4^2<14<5^2</math> is clearly false, so we discard this case.
 +
 +
The pair <math>(5,10)</math> gives <math>a=2(5)+1=11</math> and <math>n=11^2-(2(10)+1)=100,</math> which is a perfect square and therefore can be discarded.
 +
 +
The pair <math>(8,15)</math> gives <math>a=2(8)+1=17</math> and <math>n=17^2-(2(15)+1)=258,</math> which is between <math>16^2</math> and <math>17^2</math> so it is our smallest solution.
 +
 +
So, <math>\boxed{258}</math> is the correct answer.
 +
 +
~mc21s
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/tRVe2bKwIY8
 
https://youtu.be/tRVe2bKwIY8
 +
~Mathematical Dexterity
  
==See also==
+
==See Also==
 
{{AIME box|year=2021|n=II|num-b=14|after=Last Question}}
 
{{AIME box|year=2021|n=II|num-b=14|after=Last Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:30, 13 August 2021

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 $k^2<n<(k+1)^2$ 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 $k^2<n \leq (k+1)^2$.

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 $f(n)<g(n)$, 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 $k^2<n\leq(k+1)^2$ 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 $k^2<n<(k+2)^2,$ or \[n=(k+2)^2-2q\hspace{38.25mm}(3)\] for some integer $q$ such that $0<q<2k+2.$ 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*}

  • Answer

    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 $0<q<2k+2,$ respectively. Substituting $(6)$ into either inequality gives $p+1<q.$ Combining all these results produces \[0<p+1<q<2k+2. \hspace{28mm}(8)\] 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 $0<k<2m+1$. 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

See Also

2021 AIME II (ProblemsAnswer KeyResources)
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