Difference between revisions of "2005 Indonesia MO Problems/Problem 6"
Rockmanex3 (talk | contribs) (Solution to Problem 6 — three equation system) |
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x &= 1 | x &= 1 | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Therefore, one solution is <math>(1,0,-1)</math>. Since any permutation is also a solution, the solutions are <math>\boxed{(1,0,-1), (1,-1,0), (0,1,-1), (0,-1,1), (-1,0,1), (-1,1,0)} | + | Therefore, one solution is <math>(1,0,-1)</math>, and plugging that value back in satisfies the system. Since any permutation is also a solution, the solutions are <math>\boxed{(1,0,-1), (1,-1,0), (0,1,-1), (0,-1,1), (-1,0,1), (-1,1,0)}</math>, and plugging all the values back into the original system satisfies the system. |
==See Also== | ==See Also== |
Latest revision as of 10:56, 17 March 2020
Problem
Find all triples of integers which satisfy
.
Solution
By using the Distributive Property on all three equations, we get Add all three equations, rearrange, and factor to get Notice that since all the variables are integers, the squares must also be integers. The only combination of three integer squares that sum to six is .
Also note that the system is symmetric, so any permutation of a solution will work. Therefore, we can, WLOG, let .
Therefore, , , and , so , , and . By rearrangement, and .
Substitute and solve to get
Therefore, one solution is , and plugging that value back in satisfies the system. Since any permutation is also a solution, the solutions are , and plugging all the values back into the original system satisfies the system.
See Also
2005 Indonesia MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 7 |
All Indonesia MO Problems and Solutions |