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>. | ||
| + | ..... | ||
Revision as of 10:24, 20 December 2015
By expanding we get ![\[m^3+mn+m^2n^2 +n^3=m^3-3m^2n+3mb^2-n^3\]](http://latex.artofproblemsolving.com/0/d/1/0d1544ffb34dff29d0aa0c0d362cbe0b13b1a430.png)
From this the two 
cancel and you get:
![\[2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0\]](http://latex.artofproblemsolving.com/4/2/8/4280b7e66e77cbcafd88221eac9fe0f393490077.png)
We can divide by 
(nonzero).
We get:
![\[2n^2+m+m^2n+3m^2-3mn=0\]](http://latex.artofproblemsolving.com/9/8/9/98978fc9b2412fa469ad1a9a34ac576da4625162.png)
We can now factor the equation into:
![\[2n^2+(m^2-3m)n+(3m^2+m)=0\]](http://latex.artofproblemsolving.com/9/7/5/975eb800f6268d629e943923140e0a35c9af5afe.png)
From here We get the discriminant:
![\[m^4-6m^3-15m^2-8m\]](http://latex.artofproblemsolving.com/4/7/7/477f88c97356f9a4823ab4f1728617f1749383b1.png)
We factor:
![\[m(m+1)^2(m-8)\]](http://latex.artofproblemsolving.com/2/5/2/252369d13d65c2e3b9225fe223aea70cf66d0127.png)
Now we relize 
is a perfect square so 
has to be one too.
cannot be within 
mecouse that makes less then 
.
.....
