2017 USAJMO Problems/Problem 2

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 1 (and motivation)

We have $(3x^3+xy^2)(yx^2+3y^3)=(x-y)^7$ , which can be expressed as $xy(3x^2+y^2)(x^2+3y^2)=(x-y)^7$ . At this point, we think of substitution. A substitution of form $a=x+y, b=x-y$ is slightly derailed by the leftover x and y terms, so instead, seeing the xy in front, we substitute $x=a+b, y=a-b$ . This leaves us with $(a^2-b^2)(4a^2+4ab+4b^2)(4a^2-4ab+4b^2)=128b^7$ , so $(a^2-b^2)(a^2+ab+b^2)(a^2-ab+b^2)=8b^7$ . Expanding yields $a^6-b^6=8b^7$ . Rearranging, we have $b^6(8b+1)=a^6$ . To satisfy this equation in integers, $8b+1$ must obviously be a $6th$ power, and further inspection shows that it must also be odd. Also, since it is a square and all odd squares are 1 mod 8, every odd sixth power gives a solution. Since the problem asks for positive integers, the pair $(a,b)=(0,0)$ does not work. We go to the next highest odd $6th$ power, $3^6$ or $729$ . In this case, $b=91$ , so the LHS is $91^6*3^6=273^6$ , so $a=273$ . Using the original substitution yields $(x,y)=(364,182)$ as the first solution. We have shown part a by showing that there are an infinite number of positive integer solutions for $(a,b)$ , which can then be manipulated into solutions for $(x,y)$ . To solve part b, we look back at the original method of generating solutions. Define $a_n$ and $b_n$ to be the pair representing the nth solution. Since the nth odd number is $2n+1$ , $b_n=\frac{(2n+1)^6-1}{8}$ . It follows that $a_n=(2n+1)b_n=\frac{(2n+1)^7-(2n+1)}{8}$ . From our original substitution, $(x,y)=(\frac{(2n+1)^7+(2n+1)^6-2n-2}{8}, \frac{(2n+1)^7-(2n+1)^6-2n}{8})$ . {plshalp} The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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

Page

Toolbox

Search

2017 USAJMO Problems/Problem 2

Problem:

Solution 1 (and motivation)

See also