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Difference between revisions of "1994 USAMO Problems/Problem 1"
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==Solution==
 
==Solution==
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We want to show that the distance between <math>s_n</math> and <math>s_{n+1}</math> is greater than the distance between <math>s_n</math> and the next perfect square following <math>s_n</math>.
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 +
Given <math>s_n=\sum_{i=1}^{n}k_i</math>, where no <math>k_i</math> are consecutive, we can put a lower bound on <math>k_n</math>. This occurs when all <math>k_{i+1}=k_i+2</math>:
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 +
<cmath>\begin{align*}
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s_n&=(k_{n,min})+(k_{n,min}-2)+(k_{n,min}-4)+\dots+(k_{n,min}-2n+2)\\
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 +
&=nk_{n,min}-\sum_{i=1}^{n-1}2i\\
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 +
&=nk_{n,min}-2\sum_{i=1}^{n}i+2n\\
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&=nk_{n,min}-n(n+1)+2n\\
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&=nk_{n,min}-n^2+n
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\end{align*}</cmath>
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Rearranging, <math>k_{n,min}=\frac{s_n}{n}+n-1</math>. So, <math>k_n\geq\frac{s_n}{n}+n-1</math>, and the distance between <math>s_n</math> and <math>s_{n+1}</math> is <math>k_{n+1}\geq k_n+2\geq\frac{s_n}{n}+n+1</math>.
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 +
Also, let <math>d(s_n)</math> be the distance between <math>s_n</math> and the next perfect square following <math>s_n</math>. Let's look at the function <math>d(x)</math> for all positive integers <math>x</math>.
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When <math>x</math> is a perfect square, it is easy to see that <math>d(x)=2\sqrt{x}+1</math>.
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Proof: Choose <math>x=m^2</math>. <math>d(m^2)=(m+1)^2-m^2=2m+1=2\sqrt{m^2}+1</math>.
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When <math>x</math> is not a perfect square, <math>d(x)<2\sqrt{x}+1</math>.
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Proof: Choose <math>x=m^2+p</math> with <math>0<p<2m+1</math>. <math>d(m^2+p)=(m+1)^2-m^2-p=2m+1-p<2m+1=2\sqrt{m^2}+1<2\sqrt{m^2+p}+1</math>.
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So, <math>d(x)\leq 2\sqrt{x}+1</math> for all <math>x</math> and <math>d(s_n)\leq 2\sqrt{s_n}+1</math> for all <math>s_n</math>.
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Now, it suffices to show that <math>k_{n+1}\geq d(s_n)</math> for all <math>n</math>.
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<cmath>\begin{align*}
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k_{n+1}-d(s_n)&\geq \frac{s_n}{n}+n+1-2\sqrt{s_n}-1\\
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&=\frac{1}{n}(s_n+n^2-2n\sqrt{s_n})\\
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&=\frac{s_n^2+n^4+2n^2s_n-4n^2s_n}{n(s_n+n^2+2n\sqrt{s_n})}\\
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&=\frac{(s_n-n^2)^2}{n(s_n+n^2+2n\sqrt{s_n})}\\
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&\geq 0
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\end{align*}</cmath>
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So, <math>k_{n+1}\geq d(s_n)</math> and all intervals between <math>s_n</math> and <math>s_{n+1}</math> will contain at least one perfect square.
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{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:04, 29 May 2014

Let $\, k_1 < k_2 < k_3 <\cdots\,$, be positive integers, no two consecutive, and let $\, s_m = k_1+k_2+\cdots+k_m\,$, for $\, m = 1,2,3,\ldots\;\;$. Prove that, for each positive integer $n$, the interval $\, [s_n, s_{n+1})\,$, contains at least one perfect square.

Solution

We want to show that the distance between $s_n$ and $s_{n+1}$ is greater than the distance between $s_n$ and the next perfect square following $s_n$.

Given $s_n=\sum_{i=1}^{n}k_i$, where no $k_i$ are consecutive, we can put a lower bound on $k_n$. This occurs when all $k_{i+1}=k_i+2$: 

\begin{align*}
s_n&=(k_{n,min})+(k_{n,min}-2)+(k_{n,min}-4)+\dots+(k_{n,min}-2n+2)\\

&=nk_{n,min}-\sum_{i=1}^{n-1}2i\\

&=nk_{n,min}-2\sum_{i=1}^{n}i+2n\\

&=nk_{n,min}-n(n+1)+2n\\

&=nk_{n,min}-n^2+n
\end{align*} (Error compiling LaTeX. Unknown error_msg)

Rearranging, $k_{n,min}=\frac{s_n}{n}+n-1$. So, $k_n\geq\frac{s_n}{n}+n-1$, and the distance between $s_n$ and $s_{n+1}$ is $k_{n+1}\geq k_n+2\geq\frac{s_n}{n}+n+1$.

Also, let $d(s_n)$ be the distance between $s_n$ and the next perfect square following $s_n$. Let's look at the function $d(x)$ for all positive integers $x$.

When $x$ is a perfect square, it is easy to see that $d(x)=2\sqrt{x}+1$. Proof: Choose $x=m^2$. $d(m^2)=(m+1)^2-m^2=2m+1=2\sqrt{m^2}+1$.

When $x$ is not a perfect square, $d(x)<2\sqrt{x}+1$. Proof: Choose $x=m^2+p$ with $0<p<2m+1$. $d(m^2+p)=(m+1)^2-m^2-p=2m+1-p<2m+1=2\sqrt{m^2}+1<2\sqrt{m^2+p}+1$.

So, $d(x)\leq 2\sqrt{x}+1$ for all $x$ and $d(s_n)\leq 2\sqrt{s_n}+1$ for all $s_n$.

Now, it suffices to show that $k_{n+1}\geq d(s_n)$ for all $n$. 

\begin{align*}
k_{n+1}-d(s_n)&\geq \frac{s_n}{n}+n+1-2\sqrt{s_n}-1\\

&=\frac{1}{n}(s_n+n^2-2n\sqrt{s_n})\\

&=\frac{s_n^2+n^4+2n^2s_n-4n^2s_n}{n(s_n+n^2+2n\sqrt{s_n})}\\

&=\frac{(s_n-n^2)^2}{n(s_n+n^2+2n\sqrt{s_n})}\\

&\geq 0
\end{align*} (Error compiling LaTeX. Unknown error_msg)

So, $k_{n+1}\geq d(s_n)$ and all intervals between $s_n$ and $s_{n+1}$ will contain at least one perfect square.


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