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Difference between revisions of "1993 IMO Problems/Problem 1"

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==Problem==
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Let <math>f\left(x\right)=x^n+5x^{n-1}+3</math>, where <math>n>1</math> is an integer. Prove that <math>f\left(x\right)</math> cannot be expressed as the product of two non-constant polynomials with integer coefficients.
 
Let <math>f\left(x\right)=x^n+5x^{n-1}+3</math>, where <math>n>1</math> is an integer. Prove that <math>f\left(x\right)</math> cannot be expressed as the product of two non-constant polynomials with integer coefficients.
  
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For the sake of contradiction, assume that <math>f\left(x\right)=g\left(x\right)h\left(x\right)</math> for polynomials <math>g\left(x\right)</math> and <math>h\left(x\right)</math> in <math>\mathbb{R}</math>. Furthermore, let <math>g\left(x\right)=b_mx^m+b_{m-1}x^{m-1}+\ldots+b_1x+b_0</math> with <math>b_i=0</math> if <math>i>m</math> and <math>h\left(x\right)=c_{n-m}x^{n-m}+c_{n-m-1}x^{n-m-1}+\ldots+c_1x+c_0</math> with <math>c_i=0</math> if <math>i>n-m</math>. This gives that <math>f\left(x\right)=\sum_{i=0}^{n}\left(\sum_{j=0}^{i}b_jc_{i-j}\right)x^i</math>.
 
For the sake of contradiction, assume that <math>f\left(x\right)=g\left(x\right)h\left(x\right)</math> for polynomials <math>g\left(x\right)</math> and <math>h\left(x\right)</math> in <math>\mathbb{R}</math>. Furthermore, let <math>g\left(x\right)=b_mx^m+b_{m-1}x^{m-1}+\ldots+b_1x+b_0</math> with <math>b_i=0</math> if <math>i>m</math> and <math>h\left(x\right)=c_{n-m}x^{n-m}+c_{n-m-1}x^{n-m-1}+\ldots+c_1x+c_0</math> with <math>c_i=0</math> if <math>i>n-m</math>. This gives that <math>f\left(x\right)=\sum_{i=0}^{n}\left(\sum_{j=0}^{i}b_jc_{i-j}\right)x^i</math>.
  
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We have that <math>3=b_0c_0</math>, or <math>3|b_0c_0</math>. WLOG, let <math>3|b_0</math> (and thus <math>3\not|c_0</math>). Since <math>b_0c_1+b_1c_0=0</math> and <math>3</math> divides <math>b_0</math> but not <math>c_0</math>, we need that <math>3|b_1</math>. We can keep on going up the chain until we get that <math>3|b_{n-2}</math>. Then, by equating coefficients once more, we get that <math>b_0c_{n-1}+b_1c_{n-2}+\ldots+b_{n-2}c_1+b_{n-1}c_0=5</math>. Taking the equation <math>\pmod5</math> gives that <math>b_{n-1}c_0\equiv2\pmod5</math>. This implies that <math>b_{n-1}\neq0</math>. Thus, the degree of <math>g\left(x\right)</math> is at least <math>n-1</math>. However, because <math>g\left(x\right)</math> is a non-constant factor of <math>f\left(x\right)</math>, we have that the degree of <math>g\left(x\right)</math> is at most <math>n-1</math>. Thus, the degree of <math>g\left(x\right)</math> is <math>n-1</math>, implying that the degree of <math>h\left(x\right)</math> is <math>1</math>.
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We have that <math>3=b_0c_0</math>, or <math>3|b_0c_0</math>. WLOG, let <math>3|b_0</math> (and thus <math>3\not|c_0</math>). Since <math>b_0c_1+b_1c_0=0</math> and <math>3</math> divides <math>b_0</math> but not <math>c_0</math>, we need that <math>3|b_1</math>. We can keep on going up the chain until we get that <math>3|b_{n-2}</math>. Then, by equating coefficients once more, we get that <math>b_0c_{n-1}+b_1c_{n-2}+\ldots+b_{n-2}c_1+b_{n-1}c_0=5</math>. Taking the equation <math>\pmod3</math> gives that <math>b_{n-1}c_0\equiv2\pmod3</math>. This implies that <math>b_{n-1}\neq0</math>. Thus, the degree of <math>g\left(x\right)</math> is at least <math>n-1</math>. However, because <math>h\left(x\right)</math> is a non-constant factor of <math>f\left(x\right)</math>, we have that the degree of <math>g\left(x\right)</math> is at most <math>n-1</math>. Thus, the degree of <math>g\left(x\right)</math> is <math>n-1</math>, implying that the degree of <math>h\left(x\right)</math> is <math>1</math>.
  
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From this fact, we have that there must exist a rational root of <math>f\left(x\right)</math>. The only candidates are <math>1</math>, <math>-1</math>, <math>3</math>, and <math>-3</math>, though. <math>f\left(1\right)=9\neq0</math>, <math>f\left(-1\right)=\pm4+3\neq0</math>, <math>f\left(3\right)=3^n+5\cdot3^{n-1}+3>0</math>, and <math>f\left(-3\right)=\pm\left(2\cdot3^n\right)+3\neq0</math>, so none of these work. Thus, there are no linear factors of <math>f\left(x\right)</math>.
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From this fact, we have that there must exist an integer root of <math>f\left(x\right)</math>. However, when <math>x</math> is an integer, <math>x^n + 5x^{n - 1} \equiv x^{n - 2}x(x + 1) \equiv 0 \pmod{2}</math>, meaning <math>f(x)</math> is odd, so <math>x</math> cannot be a root of <math>f</math>.
  
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In other words, <math>f\left(x\right)</math> cannot be expressed as <math>g\left(x\right)h\left(x\right)</math> for polynomials <math>g\left(x\right)</math> and <math>h\left(x\right)</math> in <math>\mathbb{R}</math>. This means that <math>f\left(x\right)</math> cannot be expressed as the product of two non-constant polynomials with integer coefficients.
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Hence, <math>f\left(x\right)</math> cannot be expressed as <math>g\left(x\right)h\left(x\right)</math> for polynomials <math>g\left(x\right)</math> and <math>h\left(x\right)</math> in <math>\mathbb{R}</math>. This means that <math>f\left(x\right)</math> cannot be expressed as the product of two non-constant polynomials with integer coefficients.
  
 
Q.E.D.
 
Q.E.D.
  
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== Alternate Solution ==
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==Alternate Solution==
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It’s actually trivial by Perron’s Criterion lol
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Note: Quoting Perron's Criterion on the actual IMO will very likely result in a score in the set <math>\{0,1\}</math>, since it was not a well-known result back then.
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==See Also==
  
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Trivial by Perron's Criterion lol
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{{IMO box|year=1993|before=First Question|num-a=2}}
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(Note: Quoting Perron's Criterion on the actual IMO will very likely result in a score in the set <math>{0,1}</math>.)
 

Latest revision as of 06:08, 22 June 2024

Problem

Let $f\left(x\right)=x^n+5x^{n-1}+3$, where $n>1$ is an integer. Prove that $f\left(x\right)$ cannot be expressed as the product of two non-constant polynomials with integer coefficients.

Solution

For the sake of contradiction, assume that $f\left(x\right)=g\left(x\right)h\left(x\right)$ for polynomials $g\left(x\right)$ and $h\left(x\right)$ in $\mathbb{R}$. Furthermore, let $g\left(x\right)=b_mx^m+b_{m-1}x^{m-1}+\ldots+b_1x+b_0$ with $b_i=0$ if $i>m$ and $h\left(x\right)=c_{n-m}x^{n-m}+c_{n-m-1}x^{n-m-1}+\ldots+c_1x+c_0$ with $c_i=0$ if $i>n-m$. This gives that $f\left(x\right)=\sum_{i=0}^{n}\left(\sum_{j=0}^{i}b_jc_{i-j}\right)x^i$.

We have that $3=b_0c_0$, or $3|b_0c_0$. WLOG, let $3|b_0$ (and thus $3\not|c_0$). Since $b_0c_1+b_1c_0=0$ and $3$ divides $b_0$ but not $c_0$, we need that $3|b_1$. We can keep on going up the chain until we get that $3|b_{n-2}$. Then, by equating coefficients once more, we get that $b_0c_{n-1}+b_1c_{n-2}+\ldots+b_{n-2}c_1+b_{n-1}c_0=5$. Taking the equation $\pmod3$ gives that $b_{n-1}c_0\equiv2\pmod3$. This implies that $b_{n-1}\neq0$. Thus, the degree of $g\left(x\right)$ is at least $n-1$. However, because $h\left(x\right)$ is a non-constant factor of $f\left(x\right)$, we have that the degree of $g\left(x\right)$ is at most $n-1$. Thus, the degree of $g\left(x\right)$ is $n-1$, implying that the degree of $h\left(x\right)$ is $1$.

From this fact, we have that there must exist an integer root of $f\left(x\right)$. However, when $x$ is an integer, $x^n + 5x^{n - 1} \equiv x^{n - 2}x(x + 1) \equiv 0 \pmod{2}$, meaning $f(x)$ is odd, so $x$ cannot be a root of $f$.

Hence, $f\left(x\right)$ cannot be expressed as $g\left(x\right)h\left(x\right)$ for polynomials $g\left(x\right)$ and $h\left(x\right)$ in $\mathbb{R}$. This means that $f\left(x\right)$ cannot be expressed as the product of two non-constant polynomials with integer coefficients.

Q.E.D.

Alternate Solution

It’s actually trivial by Perron’s Criterion lol


Note: Quoting Perron's Criterion on the actual IMO will very likely result in a score in the set $\{0,1\}$, since it was not a well-known result back then.

See Also

1993 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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