1985 OIM Problems/Problem 1

Revision as of 22:28, 8 April 2024 by Clarkculus (talk | contribs) (Solution)

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 $\boxed{permutations of (5, 8, 11)}$.

See also

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