Difference between revisions of "2001 IMO Problems/Problem 6"
(→Problem 6) |
|||
(6 intermediate revisions by 4 users not shown) | |||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
− | + | ||
− | + | ||
+ | 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>. | ||
Similarly, <math>(KM+LN)-(KN+LM)=(K-L)(M-N)>0</math> since <math>K>L</math> and <math>M>N</math>. Thus, <math>KM+LN>KN+LM</math>. | Similarly, <math>(KM+LN)-(KN+LM)=(K-L)(M-N)>0</math> since <math>K>L</math> and <math>M>N</math>. Thus, <math>KM+LN>KN+LM</math>. | ||
Line 12: | Line 13: | ||
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 | + | <cmath>L^2+LN+N^2=K^2-KM+M^2</cmath> |
So, we have: | So, we have: | ||
− | <cmath>(KM+LN)(L^2+LN+N^2) | + | <cmath>(KM+LN)(L^2+LN+N^2)=KM(L^2+LN+N^2)+LN(L^2+LN+N^2)</cmath> |
− | <cmath> | + | <cmath>=KM(L^2+LN+N^2)+LN(K^2-KM+M^2)</cmath> |
− | <cmath> | + | <cmath>=KML^2+KMN^2+K^2LN+LM^2N</cmath> |
− | <cmath> | + | <cmath>=(KL+MN)(KN+LM)</cmath> |
Thus, it follows that <cmath>(KM+LN) \mid (KL+MN)(KN+LM).</cmath> | Thus, it follows that <cmath>(KM+LN) \mid (KL+MN)(KN+LM).</cmath> | ||
Now, since <math>KL+MN>KM+LN</math> if <math>KL+MN</math> is prime, then there are no common factors between the two. So, in order to have <cmath>(KM+LN)\mid (KL+MN)(KN+LM),</cmath> we would have to have <cmath>(KM+LN) \mid (KN+LM).</cmath> This is impossible as <math>KM+LN>KN+LM</math>. Thus, <math>KL+MN</math> must be composite. | Now, since <math>KL+MN>KM+LN</math> if <math>KL+MN</math> is prime, then there are no common factors between the two. So, in order to have <cmath>(KM+LN)\mid (KL+MN)(KN+LM),</cmath> we would have to have <cmath>(KM+LN) \mid (KN+LM).</cmath> This is impossible as <math>KM+LN>KN+LM</math>. Thus, <math>KL+MN</math> must be composite. | ||
+ | |||
==See also== | ==See also== | ||
{{IMO box|num-b=5|num-a=6|year=2001}} | {{IMO box|num-b=5|num-a=6|year=2001}} | ||
[[Category: Olympiad Number Theory Problems]] | [[Category: Olympiad Number Theory Problems]] |
Latest revision as of 15:36, 20 December 2018
Problem 6
are positive integers such that . Prove that is not prime.
Solution
First, as and . Thus, .
Similarly, since and . Thus, .
Putting the two together, we have
Now, we have: So, we have: 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 |