Difference between revisions of "2017 USAJMO Problems/Problem 2"

(Solution)
(Solution)
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Therefore, we have  
 
Therefore, we have  
 
<cmath>a = \left(3(n+1)^3 + (n+1)n^2 \right) \left(3n^3 + n(n+1)^2 \right),</cmath>
 
<cmath>a = \left(3(n+1)^3 + (n+1)n^2 \right) \left(3n^3 + n(n+1)^2 \right),</cmath>
which implies that there is a solution for every positive integer <math>n</math>
+
which implies that there is a solution for every positive integer <math>n</math>.
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:15, 19 April 2017

Problem:

Consider the equation \[\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.\]

(a) Prove that there are infinitely many pairs $(x,y)$ of positive integers satisfying the equation.

(b) Describe all pairs $(x,y)$ of positive integers satisfying the equation.

Solution

Part a: Let $y = an$ and $x = a(n + 1)$. Substituting, we have \[a^7 = a^6 \left(3(n+1)^3 + (n+1)n^2 \right) \left(3n^3 + n(n+1)^2 \right).\] Therefore, we have \[a = \left(3(n+1)^3 + (n+1)n^2 \right) \left(3n^3 + n(n+1)^2 \right),\] which implies that there is a solution for every positive integer $n$.

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See also

2017 USAJMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAJMO Problems and Solutions
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