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

Revision as of 10:24, 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 $0$ .

@@ Line 1: / Line 1: @@
-By expanding you get <cmath>m^3+mn+m^2n^2 +n^3=m^3-3m^2n+3mb^2-n^3</cmath>
+By expanding we get <cmath>m^3+mn+m^2n^2 +n^3=m^3-3m^2n+3mb^2-n^3</cmath>
 From this the two <math>m^3</math> cancel and you get:
 <cmath>2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0</cmath>
-You can divide by <math>n</math> (nonzero).
+We can divide by <math>n</math> (nonzero).
-You get:
+We get:
 <cmath>2n^2+m+m^2n+3m^2-3mn=0</cmath>
-You can now factor  the equation into:
+We can now factor  the equation into:
 <cmath>2n^2+(m^2-3m)n+(3m^2+m)=0</cmath>
+From here We get the discriminant:
+<cmath>m^4-6m^3-15m^2-8m</cmath>
+We factor:
+<cmath>m(m+1)^2(m-8)</cmath>
+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>.

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

Revision as of 10:24, 20 December 2015