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Difference between revisions of "1987 USAMO Problems/Problem 1"
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==Solution==
 
==Solution==
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Simply both sides completely <cmath>m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3</cmath>
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Expanding both sides, <cmath>m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3</cmath>
−
Canceling out like terms gives us <cmath>mn+m^2n^2+n^3=-3m^2n+3mn^2-n^3</cmath>
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Note that <math>m^3</math> can be canceled and as <math>n \neq 0</math>, <math>n</math> can be factored out.
−
Moving <math>2n^3</math> to the RHS and factoring out the <math>3n</math> gives us <cmath>n(m+m^2n-n^2)=3n(-m^2+mn-n^2)</cmath>
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Writing this as a quadratic equation in <math>n</math>: <cmath>2n^2+(m^2-3m)n+(3m^2+m)=0</cmath>.
−
Only nonzero solutions are needed, so <math>n</math> can be divided off. <cmath>m+m^2n-n^2=3(-m^2+mn-n^2)</cmath>
+
The discriminant <math>b^2-4ac</math> equals <cmath>(m^2-3m)^2-8(3m^2+m)</cmath>
−
Move all terms with factors of <math>n</math> to the RHS and simplifying <cmath>m-3m^2=3mn-2n^2-m^2n</cmath>
+
<cmath>=m^4-6m^3-15m^2-8m</cmath>, which we want to be a perfect square.
−
We should remove the cubic term. .......
+
Miraculously, this factors as <math>m(m-8)(m+1)^2</math>. This is square iff <math>m^2-8m</math> is square. It can be checked that the only nonzero <math>m</math> that work are <math>-1, 8, 9</math>. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs <math>(m, n)</math> as <cmath>\{(-1,-1),(8,-10),(9,-6),(9,-21)\}</cmath>.
 +
 
 
==See Also==
 
==See Also==
  

Revision as of 19:57, 10 April 2018

Problem

Find all solutions to $(m^2+n)(m + n^2)= (m - n)^3$, where m and n are non-zero integers.

Solution

Expanding both sides, \[m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3\] Note that $m^3$ can be canceled and as $n \neq 0$, $n$ can be factored out. Writing this as a quadratic equation in $n$: \[2n^2+(m^2-3m)n+(3m^2+m)=0\]. The discriminant $b^2-4ac$ equals \[(m^2-3m)^2-8(3m^2+m)\] \[=m^4-6m^3-15m^2-8m\], which we want to be a perfect square. Miraculously, this factors as $m(m-8)(m+1)^2$. This is square iff $m^2-8m$ is square. It can be checked that the only nonzero $m$ that work are $-1, 8, 9$. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs $(m, n)$ as \[\{(-1,-1),(8,-10),(9,-6),(9,-21)\}\].

See Also

1987 USAMO (Problems • Resources)
Preceded by
First
Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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