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Difference between revisions of "1987 USAMO Problems/Problem 1"
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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>
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==Problem==
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From this the two <math>m^3</math> cancel and you get:
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Find all solutions to <math>(m^2+n)(m + n^2)= (m - n)^3</math>, where m and n are non-zero integers.
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<cmath>2n^3 +mn+m^2n^2 + 3m^2n - 3mn^2=0</cmath>
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We can divide by <math>n</math> (nonzero).
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Do it
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We get:
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<cmath>2n^2+m+m^2n+3m^2-3mn=0</cmath>
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==Solution==
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We can now factor  the equation into:
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Expanding both sides, <cmath>m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3</cmath>
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<cmath>2n^2+(m^2-3m)n+(3m^2+m)=0</cmath>
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Note that <math>m^3</math> can be canceled and as <math>n \neq 0</math>, <math>n</math> can be factored out.
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From here We get the discriminant:
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Writing this as a quadratic equation in <math>n</math>: <cmath>2n^2+(m^2-3m)n+(3m^2+m)=0</cmath>.
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<cmath>m^4-6m^3-15m^2-8m</cmath>
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The discriminant <math>b^2-4ac</math> equals <cmath>(m^2-3m)^2-8(3m^2+m)</cmath>
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We factor:
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<cmath>=m^4-6m^3-15m^2-8m</cmath>, which we want to be a perfect square.
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<cmath>m(m+1)^2(m-8)</cmath>
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Miraculously, this factors as <math>m(m-8)(m+1)^2</math>. This is square iff (if and only if) <math>m^2-8m</math> is square or <math>m+1=0</math>. It can be checked that the only nonzero <math>m</math> that work are <math>-1, 8, 9</math>. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs <math>(m, n)</math> as <cmath>\{(-1,-1),(8,-10),(9,-6),(9,-21)\}</cmath>.
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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.
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<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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==See Also==
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.....
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{{USAMO box|year=1987|before=First<br>Problem|num-a=2}}
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{{MAA Notice}}
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[[Category:Olympiad Number Theory Problems]]

Latest revision as of 01:27, 11 May 2024

Problem

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

Do it

Solution

Expanding both sides, \[m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3\] Note that $m^3$ can be canceled and as $n \neq 0$, $n$ can be factored out. Writing this as a quadratic equation in $n$: \[2n^2+(m^2-3m)n+(3m^2+m)=0\]. The discriminant $b^2-4ac$ equals \[(m^2-3m)^2-8(3m^2+m)\] \[=m^4-6m^3-15m^2-8m\], which we want to be a perfect square. Miraculously, this factors as $m(m-8)(m+1)^2$. This is square iff (if and only if) $m^2-8m$ is square or $m+1=0$. It can be checked that the only nonzero $m$ that work are $-1, 8, 9$. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs $(m, n)$ as \[\{(-1,-1),(8,-10),(9,-6),(9,-21)\}\].

See Also

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

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