Problem:
Consider the equation
(3x3+xy2)(x2y+3y3)=(x−y)7
(1) Prove that there are infinitely many pairs (x,y) of positive integers satisfying the equation.
(2) Describe all pairs (x,y) of positive integers satisfying the equation.
Solution:
For x>0 and y>0, the left-hand side of the equation is positive, implying that x>y.
(1) Set yx=k+1, for some positive rational number k. Then the equation is equivalent to
(k+1)(3k2+6k+4)(k2+2k+4)=(k7)y
Take any positive integer n. Letting k=n1 yields an infinite family of solutions
(x,y)=(n(n+1)2(4n2+6n+3)(4n2+2n+1),n2(n+1)(4n2+6n+3)(4n2+2n+1))
to the given equation.
(2) Write the equation as
x(3x2+y2)y(x2+3y2)=(x−y)7
which is equivalent to
(x3+3xy2)(3x2y+y3)=(x−y)7
Let x3+3xy2=a and 3x2y+y3=b. Then a+b=(x+y)3,a−b=(x−y)3 and the equation becomes
(ab)3=(a−b)7
Let d=gcd(a,b). Then a=du and b=dv for some relatively prime positive integers u and v. Hence
(uv)3=d(u−v)7
Because gcd(u,v)=1, we have gcd(u−v,u)=1,gcd(u−v,v)=1, hence gcd(u−v,uv)=1. It follows that u−v=1 and d=(uv)3. Hence u=k+1 and v=k, where k is a positive integer, and so a=(k+1)4k3 and b=k4(k+1)3. Then
(x−y)3=a−b=[k(k+1)]3
and
(x+y)3=a+b=[k(k+1)]3(2k+1)
It follows that 2k+1=n3 for some odd integer n>1 and that x+y=nk(k+1) and x−y=k(k+1). Hence
(x,y)=(2(n+1)k(k+1),2(n−1)k(k+1))
where k=2n3−1. Thus
(x,y)=(8(n+1)(n6−1),8(n−1)(n6−1))
where n is an odd integer greater than 1 , and it is easy to check that these are solutions to the given equation. Hence these pairs describe all the solutions to the equation.
The problems on this page are the property of the MAA's American Mathematics Competitions