Difference between revisions of "1982 IMO Problems/Problem 4"
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Prove that if <math>n</math> is a positive integer such that the equation <math>x^3-3xy^2+y^3=n</math> has a solution in integers <math>x,y</math>, then it has at least three such solutions. Show that the equation has no solutions in integers for <math>n=2891</math>. | Prove that if <math>n</math> is a positive integer such that the equation <math>x^3-3xy^2+y^3=n</math> has a solution in integers <math>x,y</math>, then it has at least three such solutions. Show that the equation has no solutions in integers for <math>n=2891</math>. | ||
+ | ==Solution== | ||
Suppose the equation <math>x^3-3xy^2+y^3=n</math> has solution in integers <math>(x,y)</math> with <math>y=x+k</math>. Then, completing the cube yields <math>(y-x)^3-3x^2y+2x^3</math>. Using the substitution <math>y=x+k</math> yields <math>k^3-3kx^2-x^3=n</math>. Notice that equality directly implies that <math>(k,-x)</math> is also a solution to the original equation. Applying the transformation again yields that <math>(-x-k,-k)</math> is also a solution. We show that these three solutions are indeed distinct: If <math>(x,y)=(k,-x)</math> then <math>x=k,x+k=-x</math> which only has solution <math>x=k=0</math> which implies that <math>n</math> is not a positive integer, a contradiction. Similarly, since the transformation from <math>(k,-x)</math> to <math>(-x-k,-k)</math> and <math>(-x-k,-k)</math> to <math>(x,y)</math> is the same as the transformation from <math>(x,y)</math> to <math>(k,-x)</math>, we have that the three solutions are pairwise distinct. | Suppose the equation <math>x^3-3xy^2+y^3=n</math> has solution in integers <math>(x,y)</math> with <math>y=x+k</math>. Then, completing the cube yields <math>(y-x)^3-3x^2y+2x^3</math>. Using the substitution <math>y=x+k</math> yields <math>k^3-3kx^2-x^3=n</math>. Notice that equality directly implies that <math>(k,-x)</math> is also a solution to the original equation. Applying the transformation again yields that <math>(-x-k,-k)</math> is also a solution. We show that these three solutions are indeed distinct: If <math>(x,y)=(k,-x)</math> then <math>x=k,x+k=-x</math> which only has solution <math>x=k=0</math> which implies that <math>n</math> is not a positive integer, a contradiction. Similarly, since the transformation from <math>(k,-x)</math> to <math>(-x-k,-k)</math> and <math>(-x-k,-k)</math> to <math>(x,y)</math> is the same as the transformation from <math>(x,y)</math> to <math>(k,-x)</math>, we have that the three solutions are pairwise distinct. | ||
For the case <math>n=2891</math>, notice that <math>7|n</math>. Considering all solutions modulo <math>7</math> of the equation <math>x^3-3xy^2+y^3\equiv0\pmod{7}</math> yields only <math>x\equiv y\equiv 0\pmod{7}</math>. But, this implies that <math>7^3</math> divides <math>2891</math> which is clearly not true. | For the case <math>n=2891</math>, notice that <math>7|n</math>. Considering all solutions modulo <math>7</math> of the equation <math>x^3-3xy^2+y^3\equiv0\pmod{7}</math> yields only <math>x\equiv y\equiv 0\pmod{7}</math>. But, this implies that <math>7^3</math> divides <math>2891</math> which is clearly not true. | ||
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+ | {{IMO box|year=1982|num-b=3|num-a=5}} | ||
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+ | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 21:52, 7 December 2008
Problem
Prove that if is a positive integer such that the equation has a solution in integers , then it has at least three such solutions. Show that the equation has no solutions in integers for .
Solution
Suppose the equation has solution in integers with . Then, completing the cube yields . Using the substitution yields . Notice that equality directly implies that is also a solution to the original equation. Applying the transformation again yields that is also a solution. We show that these three solutions are indeed distinct: If then which only has solution which implies that is not a positive integer, a contradiction. Similarly, since the transformation from to and to is the same as the transformation from to , we have that the three solutions are pairwise distinct.
For the case , notice that . Considering all solutions modulo of the equation yields only . But, this implies that divides which is clearly not true.
1982 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |