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Difference between revisions of "1977 USAMO Problems/Problem 1"

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<math> (1\plus{}x^n\plus{}x^{2n}\plus{}\cdots\plus{}x^{mn})</math> is divisible by <math> (1\plus{}x\plus{}x^2\plus{}\cdots\plus{}x^{m})</math>.
 
<math> (1\plus{}x^n\plus{}x^{2n}\plus{}\cdots\plus{}x^{mn})</math> is divisible by <math> (1\plus{}x\plus{}x^2\plus{}\cdots\plus{}x^{m})</math>.
  
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== (Incorrect) Solution ==
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== Solution 1 ==
  
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Denote the first and larger polynomial to be <math>f(x)</math> and the second one to be <math>g(x)</math>. In order for <math>f(x)</math> to be divisible by <math>g(x)</math> they must have the same roots. The roots of <math>g(x)</math> are the mth roots of unity, except for 1. When plugging into <math>f(x)</math>, the root of unity is a root of <math>f(x)</math> if and only if the terms <math>x^n, x^{2n}, x^{3n}, \cdots x^{mn}</math> all represent a different mth root of unity.  
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Denote the first and larger polynomial to be <math>f(x)</math> and the second one to be <math>g(x)</math>. In order for <math>f(x)</math> to be divisible by <math>g(x)</math> they must have the same roots. The roots of <math>g(x)</math> are the (m+1)th roots of unity, except for 1. When plugging into <math>f(x)</math>, the root of unity is a root of <math>f(x)</math> if and only if the terms <math>x^n, x^{2n}, x^{3n}, \cdots x^{mn}</math> all represent a different (m+1)th root of unity not equal to 1.  
  
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Note that if <math>\gcd(m,n)=1</math>, the numbers <math>n, 2n, 3n, \cdots, mn</math> represent a complete set of residues modulo <math>m</math>. Therefore, <math>f(x)</math> divides <math>g(x)</math> only if <math>\boxed{\gcd(m,n)=1}</math> <math>\blacksquare</math>
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Note that if <math>\gcd(m+1,n)=1</math>, the numbers <math>n, 2n, 3n, \cdots, mn</math> represent a complete set of residues minus 0 modulo <math>m+1</math>. However, if <math>gcd(m+1,n)=a</math> not equal to 1, then <math>\frac{(m+1)(n)}{a}</math> is congruent to <math>0 \pmod {m+1}</math> and thus a complete set is not formed. Therefore, <math>f(x)</math> divides <math>g(x)</math> if and only if <math>\boxed{\gcd(m+1,n)=1}.</math> <math>\blacksquare</math>
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'''This solution is incorrect, but why??? (Answer below)'''
 
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== Correction to Solution ==
 
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This is a common misconception: the roots of <math>1 + x + x^2 + ... + x^m</math> are the <math>(m+1)</math>th roots of unity, not <math>m</math>th roots of unity! Thus, replace all instances of <math>m</math> with <math>m+1</math> in the above solution to produce a final answer of <math>\boxed{\gcd(m+1,n)=1}</math>, and the solution should obtain full credit....
 
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Actually, there is one more thing missing. We proved that if <math>gcd(m+1,n)=1</math>, then <math>f(x)</math> is divisible by <math>g(x).</math> But we have not proved that if <math>gcd(m+1,n) = a</math> is not one, then <math>f(x)</math> is not divisible by <math>g(x)</math>. But this problem is easily remedied, for we can show that <math>x^{\frac{m+1}{a} \cdot n} = 1</math> because of the properties of gcd, and thus the terms do not all represent a different mth root of unity.
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 20:19, 14 August 2014

Problem

Determine all pairs of positive integers $(m,n)$ such that $(1\plus{}x^n\plus{}x^{2n}\plus{}\cdots\plus{}x^{mn})$ (Error compiling LaTeX. Unknown error_msg) is divisible by $(1\plus{}x\plus{}x^2\plus{}\cdots\plus{}x^{m})$ (Error compiling LaTeX. Unknown error_msg).

Solution 1

Denote the first and larger polynomial to be $f(x)$ and the second one to be $g(x)$. In order for $f(x)$ to be divisible by $g(x)$ they must have the same roots. The roots of $g(x)$ are the (m+1)th roots of unity, except for 1. When plugging into $f(x)$, the root of unity is a root of $f(x)$ if and only if the terms $x^n, x^{2n}, x^{3n}, \cdots x^{mn}$ all represent a different (m+1)th root of unity not equal to 1.

Note that if $\\gcd(m+1,n)=1$, the numbers $n, 2n, 3n, \cdots, mn$ represent a complete set of residues minus 0 modulo $m+1$. However, if $gcd(m+1,n)=a$ not equal to 1, then $\frac{(m+1)(n)}{a}$ is congruent to $0 \pmod {m+1}$ and thus a complete set is not formed. Therefore, $f(x)$ divides $g(x)$ if and only if $\boxed{\\gcd(m+1,n)=1}.$ $\blacksquare$

Solution 2

We could instead consider $f(x)$ modulo $g(x)$. Notice that $x^{m+1} = 1 \pmod {g(x)}$, and thus we can reduce the exponents of $f(x)$ to their equivalent modulo $m+1$. We want the resulting $h(x)$ with degree less than $m+1$ to be equal to $g(x)$ (of degree $m$), which implies that the exponents of $f(x)$ must be all different modulo $m+1$. This can only occur if and only if $gcd(m+1, n) = 1$, and this is our answer, as shown in Solution 1.

See Also

1977 USAMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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