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>.
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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{\text{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

Latest revision as of 22:28, 8 April 2024

Problem

Find all triples of integers $(a,b,c)$ such that: \[a+b+c=24\] \[a^2+b^2+c^2=210\] \[abc=440\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Square the first equation to get $a^2+b^2+c^2+2(ab+bc+ca)=576$, and using the second equation $ab+bc+ca=183$. Then with the elementary symmetric sums with three variables, their solutions are the solutions to the cubic $x^3-24x^2+183x-440=0$. With a little rational root bash and factoring we get $x=5,8,11$. Thus the integer solutions are the $\boxed{\text{permutations of }(5, 8, 11)}$.

See also

https://www.oma.org.ar/enunciados/ibe1.htm