# Difference between revisions of "2002 IMO Shortlist Problems/A3"

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== Problem == | == Problem == | ||

− | Let <math> | + | Let <math>P </math> be a [[cubic polynomial | cubic]] [[polynomial]] given by <math> P(x) = ax^3 + bx^2 + cx + d </math>, where <math> a, b, c, d </math> are [[integer]]s and <math> a \neq 0 </math>. Suppose that <math> xP(x) = yP(y) </math> for infinitely many pairs <math>x </math>, <math>y </math> of integers with <math> x \neq y </math>. Prove that the equation <math> P(x) = 0 </math> has an integer root. |

== Solutions == | == Solutions == | ||

− | We note that the condition <math> | + | We note that the condition <math>xP(x) = yP(y) </math> is equivalent to |

<center> | <center> | ||

<math> | <math> | ||

− | + | a(x^4 - y^4) + b(x^3 - y^3) + c(x^2 - y^2) + d(x - y) = 0 | |

</math>. | </math>. | ||

</center> | </center> | ||

− | Since <math> | + | Since <math>x \neq y </math>, we may remove the factor <math>(x-y) </math> to obtain |

<center> | <center> | ||

<math> | <math> | ||

− | + | a(x^3 + x^{2}y + xy^2 + y^3) + b(x^2 + xy + y^2) + c(x + y) + d = 0 | |

</math>, | </math>, | ||

</center> | </center> | ||

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<center> | <center> | ||

<math> | <math> | ||

− | + | P(x+y) = xy\left[ 2a(x+y) + b \right] | |

</math>. | </math>. | ||

</center> | </center> | ||

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− | We can denote <math> | + | We can denote <math>x + y </math> and <math>xy </math> as <math>s </math> and <math>t </math>, respectively, rewriting this as |

<center> | <center> | ||

<math> | <math> | ||

− | + | P(s) = (2as + b)t | |

</math>. | </math>. | ||

</center> | </center> | ||

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=== Solution 1 === | === Solution 1 === | ||

− | We claim that <math> | + | We claim that <math>s </math> can only assume finitely many values. We note that <math>{} s^2 - 4t = (x-y)^2 \ge 0 </math>, so <math>|t| < s^2 /4 </math>, which brings us to |

<center> | <center> | ||

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</center> | </center> | ||

− | which is clearly true for at most finitely many integer values of <math> | + | which is clearly true for at most finitely many integer values of <math>s </math>. |

− | We denote <math> | + | We denote <math>xP(x) </math> by <math>Q(x) </math>. The condition <math>xP(x) = yP(y) </math> is then equivalent to <math>Q(x) = Q(y) </math>, or <math>Q(s) = Q(r-s)</math>. But <math>s </math> can only assume finitely many values, so for some value of <math>s </math>, <math>Q(x) = Q(s-x) </math> are quartic polynomials equivalent at infinitely many points and are therefore equivalent. |

− | Now, if <math> | + | Now, if <math>s = 0 </math>, then we have <math>Q(x) = Q(-x)</math>, so <math>Q </math> is an even function. Since <math>Q </math> is divisible by <math>x </math>, it must therefore also be divisible by <math>x^2 </math>, implying that <math>P </math> is divisible by <math>x </math>, which means that 0 is a root of <math>P </math>, as desired. |

− | If <math> | + | If <math>s \neq 0 </math>, then we have <math>xP(x) = (s-x)P(s-x) </math>, so <math>sP(s) = 0</math>, implying that <math>P(s) </math> is equal to 0 since <math>s \neq 0</math>. Thus <math>s </math> is the desired root of <math>P(s) </math>. |

=== Solution 2 === | === Solution 2 === | ||

− | Consider the function <math> | + | Consider the function <math>xP(x) </math>. If <math>P(x) </math> has roots <math> \lambda_1 \le \lambda_2 \le \lambda_3 </math>, then <math>P(x) </math> is bounded on the interval <math> \left[ \min (\lambda_1 , 0) , \max (\lambda_3 , 0) \right] </math>, and [[injective]] on each of the intervals <math> (-\infty , \min (\lambda_1 , 0) ) , ( \max (\lambda_3 , 0 ) , \infty) </math>. Because each value of <math>xP(x) </math> yields at most finitely many solutions <math>x,y </math>, there are at most finitely many solutions <math>x,y </math> such that <math>x </math> and <math>y </math> have the same sign. |

− | Since <math> | + | Since <math>P(s) = xy( 2as + b ) </math>, if <math>P(s) \neq 0 </math>, then each value of <math>s </math> yields at most one solution. But for arbitrarily large values of <math>s </math>, <math>P(s) </math> and <math>(2as + b) </math> have the same sign, making <math>P(s) = xy (2as + b) </math> impossible if <math>x </math> and <math>y </math> have different signs. Thus there must be some integer <math>s </math> such that <math>P(s) = 0 </math>. (Indeed, if <math>P(s) = 2as + b = 0 </math>, then any pair of integers with sum <math>s </math> is a solution.) |

## Latest revision as of 18:51, 8 September 2009

## Problem

Let be a cubic polynomial given by , where are integers and . Suppose that for infinitely many pairs , of integers with . Prove that the equation has an integer root.

## Solutions

We note that the condition is equivalent to

.

Since , we may remove the factor to obtain

,

or

.

We can denote and as and , respectively, rewriting this as

.

### Solution 1

We claim that can only assume finitely many values. We note that , so , which brings us to

,

which is clearly true for at most finitely many integer values of .

We denote by . The condition is then equivalent to , or . But can only assume finitely many values, so for some value of , are quartic polynomials equivalent at infinitely many points and are therefore equivalent.

Now, if , then we have , so is an even function. Since is divisible by , it must therefore also be divisible by , implying that is divisible by , which means that 0 is a root of , as desired.

If , then we have , so , implying that is equal to 0 since . Thus is the desired root of .

### Solution 2

Consider the function . If has roots , then is bounded on the interval , and injective on each of the intervals . Because each value of yields at most finitely many solutions , there are at most finitely many solutions such that and have the same sign.

Since , if , then each value of yields at most one solution. But for arbitrarily large values of , and have the same sign, making impossible if and have different signs. Thus there must be some integer such that . (Indeed, if , then any pair of integers with sum is a solution.)

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*