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

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 $91^6\cdot3^6=273^6$ , 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, $(x,y)=\left(\frac{(2n+1)^7+(2n+1)^6-2n-2}{8}, \frac{(2n+1)^7-(2n+1)^6-2n} {8}\right)$ .

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: $(x,y)=\left(\frac{(a+1)(a^6-1)}{8},\frac{(a-1)(a^6-1)}{8}\right)$ 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

Solution $(1.5, x)$ where $|x|>0$

So named because it is a mix of solutions 1 and 2 but differs in other aspects. After fruitless searching, let $x+y=a$ , $x-y=b$ . Clearly $a, b > 0$ . Then $x=\frac{a+b}{2}, y=\frac{a-b}{2}, x^2+y^2=\frac{a^2+b^2}{2}, x^2=\frac{a^2+2ab+b^2}{4}, y^2=\frac{a^2-2ab+b^2}{4}, xy=\frac{a^2-b^2}{4}$ .

Change the LHS to $xy(3x^2+y^2)(x^2+3y^2)=(a-b)(a+b)(a^2+ab+b^2)(a^2-ab+b^2)*\frac{1}{4}=(a^3-b^3)(a^3+b^3)*\frac{1}{4}=(a^6-b^6)*\frac{1}{4}$ . Change the RHS to $b^7$ . Therefore $(\frac{a}{b})^6=4b+1$ . Let $n=\frac{a}{b}$ , and note that $n$ is an integer. Therefore $a=n(\frac{n^6-1}{4}), b=\frac{n^6-1}{4}$ . Because $n^6 \equiv 1 (\mod{4})$ , $n$ is odd and $>1$ because $b>0$ . Therefore, substituting for $x$ and $y$ we get: $x=\frac{(n^4+n^2+1)(n-1)(n+1)^2}{8}, y=\frac{(n+1)(n-1)^2(n^4+n^2+1)}{8}$ .

Simplified Evan Chen's Solution

Let $x = da$ , $y = db$ , such that $gcd(a,b) = 1$ . It's obvious $x > y$ , so $a > b$

By plugging it in to the original equation, and simplifying, we get

$d = \frac{ab(a^2 + 3b^2)(3a^2 + b^2)}{(a-b)^7}$

Since d is an integer, we got:

$(a-b)^7 \mid ab(a^2 + 3b^2)(3a^2 + b^2) \: \: \: \: \: (*)$

1. Since $gcd(a,b) = 1$ , a and b can't both be even

2. If both a and b are odd. $a-b$ will be even. The left hand side (*) has at least 7 factors of 2. Evaluating the (*) RHS term by term, we get: $a \equiv 1 \pmod{2}$ $b \equiv 1 \pmod{2}$ $a^2 + 3b^2 \equiv 4 \pmod{8}$ $3a^2 + b^2 \equiv 4 \pmod{8}$

the (*) RHS has only 4 factors of 2. Contradiction.

3. Thus, $a-b$ has to be odd. We want to prove $a-b=1$

We know: $a \equiv b \pmod{a-b}$ So $ab(a^2 + 3b^2)(3a^2 + b^2) \equiv 16a^6 \pmod{a-b}$ We can assume if $a-b \neq 1$

Since $a-b$ is odd, $gcd(16, a-b) = 1$

Since $gcd(a, a-b) = 1$ , $gcd(a^6, a-b) = 1$

$ab(a^2 + 3b^2)(3a^2 + b^2) \equiv 16a^6 \neq 0 \pmod{a-b}$ contradiction. So, $a-b = 1$ .

If $a-b=1$ , obviously (*) will work, d will be an integer.

So (*) $\Leftrightarrow$ $a-b=1$ . The proof is complete.

-Alexlikemath

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
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All USAJMO Problems and Solutions

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2017 USAJMO Problems/Problem 2

Contents

Problem:

Solution 1

Solution 1

Solution 2 (and motivation)

Solution $(1.5, x)$ where $|x|>0$

Simplified Evan Chen's Solution

See also

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Toolbox

Search

2017 USAJMO Problems/Problem 2

Contents

Problem:

Solution 1

Solution 1

Solution 2 (and motivation)

Solution where

Simplified Evan Chen's Solution

See also

Solution $(1.5, x)$ where $|x|>0$