Difference between revisions of "1985 OIM Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | { | + | Square the first equation to get <math>a^2+b^2+c^2+2(ab+bc+ca)=576</math>, and using the second equation <math>ab+bc+ca=183</math>. Then with the elementary symmetric sums with three variables, their solutions are the solutions to the cubic <math>x^3-24x^2+183x-440=0</math>. With a little rational root bash and factoring we get <math>x=5,8,11</math>. Thus the integer solutions are the <math>\boxed{permutations of (5, 8, 11)}</math>. |
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe1.htm | https://www.oma.org.ar/enunciados/ibe1.htm | ||
Revision as of 23:28, 8 April 2024
Problem
Find all triples of integers 
such that:
![\[a+b+c=24\]](http://latex.artofproblemsolving.com/5/3/4/5344af63c0d9cbb06f5387651af31a92862ad7f8.png)
![\[a^2+b^2+c^2=210\]](http://latex.artofproblemsolving.com/b/0/e/b0e6a5ee218d88785f66c9b92b5f3029bb6a1852.png)
![\[abc=440\]](http://latex.artofproblemsolving.com/c/a/b/cabbb8774fef0c2cfbb07e57ff60ba7ba937c344.png)
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Square the first equation to get 
, and using the second equation 
. Then with the elementary symmetric sums with three variables, their solutions are the solutions to the cubic 
. With a little rational root bash and factoring we get 
. Thus the integer solutions are the 
.
