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

(Adds another solution)
m (→‎Solution 2 (and motivation))
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Now, we have that <cmath>\frac{(x+y)^6-(x-y)^6}{4}=(x-y)^7</cmath>
 
Now, we have that <cmath>\frac{(x+y)^6-(x-y)^6}{4}=(x-y)^7</cmath>
 
Rearranging and getting rid of the denominator, we have that <cmath>(x+y)^6=4(x-y)^7+(x-y)^6</cmath>
 
Rearranging and getting rid of the denominator, we have that <cmath>(x+y)^6=4(x-y)^7+(x-y)^6</cmath>
−
Factoring, we have <cmath>(x+y)^6=(x-y)^6(4(x-y)+1)</cmath>Dividing both sides, we have <cmath>(\frac{x+y}{x-y})^6=4x-4y+1</cmath>
+
Factoring, we have <cmath>(x+y)^6=(x-y)^6(4(x-y)+1)</cmath>Dividing both sides, we have <cmath>\left(\frac{x+y}{x-y}\right)^6=4x-4y+1</cmath>
 
Now, since the LHS is the 6th power of a rational number, and the RHS is an integer, the RHS must be a perfect 6th power. Define <math>a=\frac{x+y}{x-y}</math>. By inspection, <math>a</math> must be a positive odd integer satistisfying <math>a \geq 3</math>. We also have <cmath>a^6=4x-4y+1</cmath> Now, we can solve for <math>x</math> and <math>y</math> in terms of <math>a</math>:
 
Now, since the LHS is the 6th power of a rational number, and the RHS is an integer, the RHS must be a perfect 6th power. Define <math>a=\frac{x+y}{x-y}</math>. By inspection, <math>a</math> must be a positive odd integer satistisfying <math>a \geq 3</math>. We also have <cmath>a^6=4x-4y+1</cmath> Now, we can solve for <math>x</math> and <math>y</math> in terms of <math>a</math>:
 
<math>x-y=\frac{a^6-1}{4}</math> and <math>x+y=a(x-y)=\frac{a(a^6-1)}{4}</math>.
 
<math>x-y=\frac{a^6-1}{4}</math> and <math>x+y=a(x-y)=\frac{a(a^6-1)}{4}</math>.

Revision as of 19:28, 31 December 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 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}

Solution 2 (and motivation)

First, we shall prove a lemma:

LEMMA:

\[\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = \frac{(x+y)^6-(x-y)^6}{4}\] PROOF: Expanding and simplifying the right side, we find that \[\frac{(x+y)^6-(x-y)^6}{4}=\frac{12x^5y+40x^3y^3+12xy^5}{4}\]\[=3x^5y+10x^3y^3+3xy^5\]\[=\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right)\] which proves our lemma.


Now, we have that \[\frac{(x+y)^6-(x-y)^6}{4}=(x-y)^7\] Rearranging and getting rid of the denominator, we have that \[(x+y)^6=4(x-y)^7+(x-y)^6\] Factoring, we have \[(x+y)^6=(x-y)^6(4(x-y)+1)\]Dividing both sides, we have \[\left(\frac{x+y}{x-y}\right)^6=4x-4y+1\] Now, since the LHS is the 6th power of a rational number, and the RHS is an integer, the RHS must be a perfect 6th power. Define $a=\frac{x+y}{x-y}$. By inspection, $a$ must be a positive odd integer satistisfying $a \geq 3$. We also have \[a^6=4x-4y+1\] Now, we can solve for $x$ and $y$ in terms of $a$: $x-y=\frac{a^6-1}{4}$ and $x+y=a(x-y)=\frac{a(a^6-1)}{4}$. Now we have: \[(x,y)=(\frac{(a+1)(a^6-1)}{8},\frac{(a-1)(a^6-1)}{8})\] and it is trivial to check that this parameterization works for all such $a$ (to keep $x$ and $y$ integral), which implies part (a).


MOTIVATION FOR LEMMA: I expanded the LHS, noticed the coefficients were $(3,10,3)$, and immediately thought of binomial coefficients. Looking at Pascal's triangle, it was then easy to find and prove the lemma.

-sunfishho

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

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