# Difference between revisions of "2015 AIME II Problems/Problem 6"

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These triples give <math>a</math> values of <math>13</math>, <math>15</math>, and <math>15</math>, respectively, and <math>c</math> values of <math>64</math>, <math>216</math>, and <math>224</math>, respectively. | These triples give <math>a</math> values of <math>13</math>, <math>15</math>, and <math>15</math>, respectively, and <math>c</math> values of <math>64</math>, <math>216</math>, and <math>224</math>, respectively. | ||

− | We know that Jon still found two possible values of <math>c</math> when Steve told him the <math>a</math> value, so the <math>a</math> value must be <math>15</math>. Thus, the two <math>c</math> values are <math>216</math> and <math>224</math>, which sum to <math>\boxed{\ | + | We know that Jon still found two possible values of <math>c</math> when Steve told him the <math>a</math> value, so the <math>a</math> value must be <math>15</math>. Thus, the two <math>c</math> values are <math>216</math> and <math>224</math>, which sum to <math>\boxed{\text{240}}</math>. |

~BealsConjecture~ | ~BealsConjecture~ |

## Revision as of 00:56, 27 March 2015

## Problem

Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form for some positive integers and . Can you tell me the values of and ?"

After some calculations, Jon says, "There is more than one such polynomial."

Steve says, "You're right. Here is the value of ." He writes down a positive integer and asks, "Can you tell me the value of ?"

Jon says, "There are still two possible values of ."

Find the sum of the two possible values of .

## Solution

We call the three roots (some may be equal to one another) , , and . Using Vieta's formulas, we get , , and .

Squaring our first equation we get .

We can then subtract twice our second equation to get .

Simplifying the right side:

So, we know .

We can then list out all the triples of positive integers whose squares sum to :

We get , , and .

These triples give values of , , and , respectively, and values of , , and , respectively.

We know that Jon still found two possible values of when Steve told him the value, so the value must be . Thus, the two values are and , which sum to .

~BealsConjecture~