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Difference between revisions of "1994 USAMO Problems/Problem 1"

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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.
 
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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==Solution 2==
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We see that by increasing <math>n</math> by some amount, we simply shift our interval by a finite amount. It suffices to consider the case <math>n=1</math> (since this can be inducted across all positive integers). Let <math>k_1=x</math>. We want the smallest interval, so we have <math>[x, 2x+2]</math>. Simple induction reveals that the ration of consecutive squares grows slower than our linear bound. We now consider sufficiently small <math>x</math> (where <math>\frac{(n+1)^2}{n^2}<2</math>). This first happens at <math>n=3</math>. By simple casework, our answer is as desired <math>\blacksquare</math>.
  
 
==See Also==
 
==See Also==

Revision as of 13:24, 22 August 2018

Problem

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*}

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*}

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.

Solution 2

We see that by increasing $n$ by some amount, we simply shift our interval by a finite amount. It suffices to consider the case $n=1$ (since this can be inducted across all positive integers). Let $k_1=x$. We want the smallest interval, so we have $[x, 2x+2]$. Simple induction reveals that the ration of consecutive squares grows slower than our linear bound. We now consider sufficiently small $x$ (where $\frac{(n+1)^2}{n^2}<2$). This first happens at $n=3$. By simple casework, our answer is as desired $\blacksquare$.

See Also

1994 USAMO (Problems • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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