Difference between revisions of "1987 USAMO Problems/Problem 1"

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==Solution==
 
==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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Canceling out like terms gives us <cmath>mn+m^2n^2+n^3=-3m^2n+3mn^2-n^3</cmath>
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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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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>
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Move all terms with factors of <math>n</math> to the RHS and simplifying <cmath>m=-3m^2+3mn-2n^2-m^2n</cmath>
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We should remove the cubic term. .......
 
==See Also==
 
==See Also==
  

Revision as of 12:27, 1 April 2017

Problem

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

Solution

Simply both sides completely \[m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3\] Canceling out like terms gives us \[mn+m^2n^2+n^3=-3m^2n+3mn^2-n^3\] Moving $2n^3$ to the RHS and factoring out the $3n$ gives us \[n(m+m^2n-n^2)=3n(-m^2+mn-n^2)\] Only nonzero solutions are needed, so $n$ can be divided off. \[m+m^2n-n^2=3(-m^2+mn-n^2)\] Move all terms with factors of $n$ to the RHS and simplifying \[m=-3m^2+3mn-2n^2-m^2n\] We should remove the cubic term. .......

See Also

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

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