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>. | + | 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 such that:
~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 .