Difference between revisions of "2001 IMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
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First, <math>(KL+MN)-(KM+LN)=(K-N)(L-M)>0</math> as <math>K>N</math> and <math>L>M</math>. Thus, <math>KL+MN>KM+LN</math>. | First, <math>(KL+MN)-(KM+LN)=(K-N)(L-M)>0</math> as <math>K>N</math> and <math>L>M</math>. Thus, <math>KL+MN>KM+LN</math>. | ||
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Now, we have: | Now, we have: | ||
− | <cmath>(K+L-M+N)(-K+L+M+N) | + | <cmath>(K+L-M+N)(-K+L+M+N)=KM+LN</cmath> |
− | <cmath>-K^2+KM+L^2+LN+KM-M^2+LN+N^2 | + | <cmath>-K^2+KM+L^2+LN+KM-M^2+LN+N^2=KM+LN</cmath> |
<cmath>L^2+LN+N^2&=&K^2-KM+M^2</cmath> | <cmath>L^2+LN+N^2&=&K^2-KM+M^2</cmath> | ||
So, we have: | So, we have: |
Revision as of 19:08, 4 December 2015
Problem 6
are positive integers such that . Prove that is not prime.
Solution
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First, as and . Thus, .
Similarly, since and . Thus, .
Putting the two together, we have
Now, we have:
\[L^2+LN+N^2&=&K^2-KM+M^2\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
So, we have:
\[(KM+LN)(L^2+LN+N^2)&=&KM(L^2+LN+N^2)+LN(L^2+LN+N^2)\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[&=&KM(L^2+LN+N^2)+LN(K^2-KM+M^2)\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[&=&KML^2+KMN^2+K^2LN+LM^2N\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[&=&(KL+MN)(KN+LM)\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
Thus, it follows that Now, since if is prime, then there are no common factors between the two. So, in order to have we would have to have This is impossible as . Thus, must be composite.
See also
2001 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |