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Difference between revisions of "2001 USAMO Problems/Problem 5"

(New page: == Problem == Let <math>S</math> be a set of integers (not necessarily positive) such that (a) there exist <math>a,b \in S</math> with <math>\gcd(a,b) = \gcd(a - 2,b - 2) = 1</math>; (b...)
 
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== Solution ==
 
== Solution ==
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{{solution}}
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In the solution below we use the expression <math>S</math> ''is stable under'' <math>x\mapsto f(x)</math> to mean that if <math>x</math> belongs to <math>S</math> then <math>f(x)</math> also belongs to <math>S</math>. If <math>c,d\in S</math>, then by (B), <math>S</math> is stable under <math>x\mapsto c^2 - x</math> and <math>x\mapsto d^2 - x</math>, hence stable under <math>x\mapsto c^2 - (d^2 - x) = x + (c^2 - d^2)</math>. Similarly <math>S</math> is stable under <math>x\mapsto x + (d^2 - c^2)</math>. Hence <math>S</math> is stable under <math>x\mapsto x + n</math> and <math>x\mapsto x - n</math> whenever <math>n</math> is an integer linear combination of numbers of the form <math>c^2 - d^2</math> with <math>c,d\in S</math>. In particular, this holds for <math>n = m</math>, where <math>m = \gcd\{c^2 - d^2 : c,d\in S\}</math>.
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Since <math>S\neq \emptyset</math> by (A), it suffices to prove that <math>m = 1</math>. For the sake of contradiction, assume that <math>m\neq 1</math>. Let <math>p</math> be a prime dividing <math>m</math>. Then <math>c^2 - d^2\equiv 0\pmod{p}</math> for all <math>c,d\in S</math>. In other words, for each <math>c,d\in S</math>, either <math>d\equiv c\pmod{p}</math> or <math>d\equiv -c\pmod{p}</math>. Given <math>c\in S</math>, <math>c^2 - c\in S</math> by (B), so <math>c^2 - c\equiv c\pmod{p}</math> or <math>c^2 - c\equiv -c\pmod{p}</math>. Hence
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<cmath>\text{For each }c\in S\text{, either }c\equiv 0\pmod{p}\text{ or }c\equiv 2\pmod{p}.\qquad\qquad (*)</cmath>
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By (A), there exist some <math>a</math> and <math>b</math> in <math>S</math> such that <math>\gcd(a,b) = 1</math>, that is, at least one of <math>a</math> or <math>b</math> cannot be divisible by <math>p</math>. Denote such an element of <math>S</math> by <math>\alpha</math>: thus, <math>\alpha\not\equiv 0\pmod{p}</math>. Similarly, by (A), <math>\gcd(a - 2, b - 2) = 1</math>, so <math>p</math> cannot divide both <math>a - 2</math> and <math>b - 2</math>. Thus, there is an element of <math>S</math>, call it <math>\beta</math>, such that <math>\beta\not\equiv 2\pmod{p}</math>. By <math>(*)</math>, <math>\alpha\equiv 2\pmod{p}</math> and <math>\beta\equiv 0\pmod{p}</math>. By (B), <math>\beta^2 - \alpha\in S</math>. Taking <math>c = \beta^2 - \alpha</math> in <math>(*)</math> yields either <math>-2\equiv 0\pmod{p}</math> or <math>-2\equiv 2\pmod{p}</math>, so <math>p = 2</math>. Now <math>(*)</math> says that all elements of <math>S</math> are even, contradicting (A). Hence our assumption is false and <math>S</math> is the set of all integers.
  
 
== See also ==
 
== See also ==
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[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 07:44, 23 October 2022

Problem

Let $S$ be a set of integers (not necessarily positive) such that

(a) there exist $a,b \in S$ with $\gcd(a,b) = \gcd(a - 2,b - 2) = 1$;

(b) if $x$ and $y$ are elements of $S$ (possibly equal), then $x^2 - y$ also belongs to $S$.

Prove that $S$ is the set of all integers.

Solution

In the solution below we use the expression $S$ is stable under $x\mapsto f(x)$ to mean that if $x$ belongs to $S$ then $f(x)$ also belongs to $S$. If $c,d\in S$, then by (B), $S$ is stable under $x\mapsto c^2 - x$ and $x\mapsto d^2 - x$, hence stable under $x\mapsto c^2 - (d^2 - x) = x + (c^2 - d^2)$. Similarly $S$ is stable under $x\mapsto x + (d^2 - c^2)$. Hence $S$ is stable under $x\mapsto x + n$ and $x\mapsto x - n$ whenever $n$ is an integer linear combination of numbers of the form $c^2 - d^2$ with $c,d\in S$. In particular, this holds for $n = m$, where $m = \gcd\{c^2 - d^2 : c,d\in S\}$.

Since $S\neq \emptyset$ by (A), it suffices to prove that $m = 1$. For the sake of contradiction, assume that $m\neq 1$. Let $p$ be a prime dividing $m$. Then $c^2 - d^2\equiv 0\pmod{p}$ for all $c,d\in S$. In other words, for each $c,d\in S$, either $d\equiv c\pmod{p}$ or $d\equiv -c\pmod{p}$. Given $c\in S$, $c^2 - c\in S$ by (B), so $c^2 - c\equiv c\pmod{p}$ or $c^2 - c\equiv -c\pmod{p}$. Hence \[\text{For each }c\in S\text{, either }c\equiv 0\pmod{p}\text{ or }c\equiv 2\pmod{p}.\qquad\qquad (*)\] By (A), there exist some $a$ and $b$ in $S$ such that $\gcd(a,b) = 1$, that is, at least one of $a$ or $b$ cannot be divisible by $p$. Denote such an element of $S$ by $\alpha$: thus, $\alpha\not\equiv 0\pmod{p}$. Similarly, by (A), $\gcd(a - 2, b - 2) = 1$, so $p$ cannot divide both $a - 2$ and $b - 2$. Thus, there is an element of $S$, call it $\beta$, such that $\beta\not\equiv 2\pmod{p}$. By $(*)$, $\alpha\equiv 2\pmod{p}$ and $\beta\equiv 0\pmod{p}$. By (B), $\beta^2 - \alpha\in S$. Taking $c = \beta^2 - \alpha$ in $(*)$ yields either $-2\equiv 0\pmod{p}$ or $-2\equiv 2\pmod{p}$, so $p = 2$. Now $(*)$ says that all elements of $S$ are even, contradicting (A). Hence our assumption is false and $S$ is the set of all integers.

See also

2001 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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