2013 China Second Round Olympiad (C) Test 2 Q3

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2013 China Second Round Olympiad (C) Test 2 Q3

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Source: 13 Oct 2013

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sqing

#1 h Oct 15, 2013, 4:10 AM • 3 Y

Y by thaygiaocht, Adventure10, Mango247

The integers $n>1$ is given . The positive integer $a_1,a_2,\cdots,a_n$ satisfing condition :
(1) $a_1<a_2<\cdots<a_n$ ;
(2) $\frac{a^2_1+a^2_2}{2},\frac{a^2_2+a^2_3}{2},\cdots,\frac{a^2_{n-1}+a^2_n}{2}$ are all perfect squares .
Prove that : $a_n\ge 2n^2-1.$

This post has been edited 1 time. Last edited by sqing, Oct 15, 2013, 7:00 AM

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#2 h Oct 15, 2013, 4:36 PM • 5 Y

Y by sqing, joybangla, dizzy, thaygiaocht, Adventure10

I will be using the known fact that if positive integers $a,b,c$ satisfy $a^2+b^2=2c^2$ then there exists integers $u$ and $v$ such that $\begin{cases}a=u^2-v^2+2uv\\b=v^2-u^2+2uv\\c=u^2+v^2\end{cases}$

For some $m\le n-1$ suppose $a_m\ge 2m^2-1$ .
Let $a_m=j^2-i^2+2ij\geq 2m^2-1$ for some positive integers $i> j$ .

We also have $i^2+j^2>a_m\geq 2m^2-1\implies i^2+j^2\geq 2m^2\implies i\geq m+1\implies i^2-j^2\geq i^2-(i-1)^2=2i-1\geq 2m+1$
Thus $a_{m+1}=i^2-j^2+2ij\geq 2m^2-1+4m+2=2(m+1)^2-1$ . Also note that $a_1\geq 2\cdot 1^2-1$ .

Hence for any integer $n>1$ we have $a_n\ge 2n^2-1$ .

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Jjesus

#3 h Apr 13, 2025, 11:03 PM

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iarnab_kundu wrote:

I will be using the known fact that if positive integers $a,b,c$ satisfy $a^2+b^2=2c^2$ then there exists integers $u$ and $v$ such that $\begin{cases}a=u^2-v^2+2uv\\b=v^2-u^2+2uv\\c=u^2+v^2\end{cases}$
For some $m\le n-1$ suppose $a_m\ge 2m^2-1$ .
Let $a_m=j^2-i^2+2ij\geq 2m^2-1$ for some positive integers $i> j$ .

We also have $i^2+j^2>a_m\geq 2m^2-1\implies i^2+j^2\geq 2m^2\implies i\geq m+1\implies i^2-j^2\geq i^2-(i-1)^2=2i-1\geq 2m+1$ Thus $a_{m+1}=i^2-j^2+2ij\geq 2m^2-1+4m+2=2(m+1)^2-1$ . Also note that $a_1\geq 2\cdot 1^2-1$ .

Hence for any integer $n>1$ we have $a_n\ge 2n^2-1$ .

$\begin{cases}a=d\cdot (u^2-v^2+2uv)\\b=d\cdot (v^2-u^2+2uv)\\c=d\cdot (u^2+v^2)\end{cases}$

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