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

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Now we relize <math>(m+1)^2</math> is a perfect square so <math>m(m-8)</math> has to be one too.
 
Now we relize <math>(m+1)^2</math> is a perfect square so <math>m(m-8)</math> has to be one too.
 
<math>m</math> cannot be within <math>0<m<8</math> mecouse that makes <math>m(m-8)</math> less then <math>0</math>.
 
<math>m</math> cannot be within <math>0<m<8</math> mecouse that makes <math>m(m-8)</math> less then <math>0</math>.
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Revision as of 11:24, 20 December 2015

By expanding we get \[m^3+mn+m^2n^2 +n^3=m^3-3m^2n+3mb^2-n^3\] From this the two $m^3$ cancel and you get: \[2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0\] We can divide by $n$ (nonzero). We get: \[2n^2+m+m^2n+3m^2-3mn=0\] We can now factor the equation into: \[2n^2+(m^2-3m)n+(3m^2+m)=0\] From here We get the discriminant: \[m^4-6m^3-15m^2-8m\] We factor: \[m(m+1)^2(m-8)\] Now we relize $(m+1)^2$ is a perfect square so $m(m-8)$ has to be one too. $m$ cannot be within $0<m<8$ mecouse that makes $m(m-8)$ less then $0$. .....