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

Revision as of 17:31, 18 July 2016

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

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-By expanding we get <cmath>m^3+mn+m^2n^2 +n^3=m^3-3m^2n+3mb^2-n^3</cmath>
+==Problem==
-From this the two <math>m^3</math> cancel and you get:
+Find all solutions to <math>(m^2+n)(m + n^2)= (m - n)^3</math>, where m and n are non-zero integers.
-<cmath>2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0</cmath>
-We can divide by <math>n</math> (nonzero).
+==Solution==
-We get:
+{{solution}}
-<cmath>2n^2+m+m^2n+3m^2-3mn=0</cmath>
-We can now factor  the equation into:
+==See Also==
-<cmath>2n^2+(m^2-3m)n+(3m^2+m)=0</cmath>
-From here We get the discriminant:
+{{USAMO box|year=1987|before=First<br>Problem|num-a=2}}
-<cmath>m^4-6m^3-15m^2-8m</cmath>
+{{MAA Notice}}
-We factor:
+[[Category:Olympiad Number Theory Problems]]
-<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>.
-.....

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