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

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(b) Describe all pairs <math>(x,y)</math> of positive integers satisfying the equation.
 
(b) Describe all pairs <math>(x,y)</math> of positive integers satisfying the equation.
  
==Solution 1 (and motivation)==
+
==Solution 1==
We have <math>(3x^3+xy^2)(yx^2+3y^3)=(x-y)^7</math>, which can be expressed as <math>xy(3x^2+y^2)(x^2+3y^2)=(x-y)^7</math>. At this point, we think of substitution. A substitution of form <math>a=x+y, b=x-y</math> is slightly derailed by the leftover x and y terms, so instead, seeing the xy in front, we substitute <math>x=a+b, y=a-b</math>. This leaves us with <math>(a^2-b^2)(4a^2+4ab+4b^2)(4a^2-4ab+4b^2)=128b^7</math>, so <math>(a^2-b^2)(a^2+ab+b^2)(a^2-ab+b^2)=8b^7</math>. Expanding yields <math>a^6-b^6=8b^7</math>. Rearranging, we have <math>b^6(8b+1)=a^6</math>. To satisfy this equation in integers, <math>8b+1</math> must obviously be a <math>6th</math> 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 <math>(a,b)=(0,0)</math> does not work. We go to the next highest odd <math>6th</math> power, <math>3^6</math> or <math>729</math>. In this case, <math>b=91</math>, so the LHS is <math>91^6*3^6=273^6</math>, so <math>a=273</math>. Using the original substitution yields <math>(x,y)=(364,182)</math> as the first solution. We have shown part a by showing that there are an infinite number of positive integer solutions for <math>(a,b)</math>, which can then be manipulated into solutions for <math>(x,y)</math>. To solve part b, we look back at the original method of generating solutions. Define <math>a_n</math> and <math>b_n</math> to be the pair representing the nth solution. Since the nth odd number is <math>2n+1</math>, <math>b_n=\frac{(2n+1)^6-1}{8}</math>. It follows that <math>a_n=(2n+1)b_n=\frac{(2n+1)^7-(2n+1)}{8}</math>. From our original substitution, <math>(x,y)=(\frac{(2n+1)^7+(2n+1)^6-2n-2}{8}, \frac{(2n+1)^7-(2n+1)^6-2n}{8})</math>.
+
We have <math>(3x^3+xy^2)(yx^2+3y^3)=(x-y)^7</math>, which can be expressed as <math>xy(3x^2+y^2)(x^2+3y^2)=(x-y)^7</math>. At this point, we  
{plshalp}
+
 
 +
think of substitution. A substitution of form <math>a=x+y, b=x-y</math> is slightly derailed by the leftover x and y terms, so  
 +
 
 +
instead, seeing the xy in front, we substitute <math>x=a+b, y=a-b</math>. This leaves us with <math>(a^2-b^2)(4a^2+4ab+4b^2)(4a^2-
 +
4ab+4b^2)=128b^7</math>, so <math>(a^2-b^2)(a^2+ab+b^2)(a^2-ab+b^2)=8b^7</math>. Expanding yields <math>a^6-b^6=8b^7</math>. Rearranging, we have  
 +
 
 +
<math>b^6(8b+1)=a^6</math>. To satisfy this equation in integers, <math>8b+1</math> must obviously be a <math>6th</math> 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 <math>(a,b)=(0,0)</math> does not work. We go to the next highest odd  
 +
 
 +
<math>6th</math> power, <math>3^6</math> or <math>729</math>. In this case, <math>b=91</math>, so the LHS is <math>91^6\cdot3^6=273^6</math>, so <math>a=273</math>. Using the original  
 +
 
 +
substitution yields <math>(x,y)=(364,182)</math> as the first solution. We have shown part a by showing that there are an infinite  
 +
 
 +
number of positive integer solutions for <math>(a,b)</math>, which can then be manipulated into solutions for <math>(x,y)</math>. To solve part  
 +
 
 +
b, we look back at the original method of generating solutions. Define <math>a_n</math> and <math>b_n</math> to be the pair representing the nth  
 +
 
 +
solution. Since the nth odd number is <math>2n+1</math>, <math>b_n=\frac{(2n+1)^6-1}{8}</math>. It follows that <math>a_n=(2n+1)b_n=\frac{(2n+1)^7-
 +
(2n+1)}{8}</math>. From our original substitution, <math>(x,y)=\left(\frac{(2n+1)^7+(2n+1)^6-2n-2}{8}, \frac{(2n+1)^7-(2n+1)^6-2n}
 +
{8}\right)</math>.
 +
 
 +
==Solution 2 (and motivation)==
 +
First, we shall prove a lemma:
 +
 
 +
LEMMA:
 +
 
 +
<cmath>\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = \frac{(x+y)^6-(x-y)^6}{4}</cmath>
 +
PROOF: Expanding and simplifying the right side, we find that <cmath>\frac{(x+y)^6-(x-y)^6}{4}=\frac{12x^5y+40x^3y^3+12xy^5}{4}</cmath><cmath>=3x^5y+10x^3y^3+3xy^5</cmath><cmath>=\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right)</cmath>
 +
which proves our lemma.
 +
 
 +
 
 +
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>
 +
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>:
 +
<math>x-y=\frac{a^6-1}{4}</math> and <math>x+y=a(x-y)=\frac{a(a^6-1)}{4}</math>.
 +
Now we have: <cmath>(x,y)=\left(\frac{(a+1)(a^6-1)}{8},\frac{(a-1)(a^6-1)}{8}\right)</cmath>
 +
and it is trivial to check that this parameterization works for all such <math>a</math> (to keep <math>x</math> and <math>y</math> integral), which implies part (a).
 +
 
 +
 
 +
MOTIVATION FOR LEMMA:
 +
I expanded the LHS, noticed the coefficients were <math>(3,10,3)</math>, and immediately thought of binomial coefficients. Looking at Pascal's triangle, it was then easy to find and prove the lemma.
 +
 
 +
-sunfishho
 +
 
 +
==Solution <math>(1.5, x)</math> where <math>|x|>0</math>==
 +
So named because it is a mix of solutions 1 and 2 but differs in other aspects.
 +
After fruitless searching, let <math>x+y=a</math>, <math>x-y=b</math>. Clearly <math>a, b > 0</math>. Then <math>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}</math>.
 +
 
 +
Change the LHS to <math>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}</math>. Change the RHS to <math>b^7</math>. Therefore <math>(\frac{a}{b})^6=4b+1</math>. Let <math>n=\frac{a}{b}</math>, and note that <math>n</math> is an integer. Therefore <math>a=n(\frac{n^6-1}{4}), b=\frac{n^6-1}{4}</math>. Because <math>n^6 \equiv 1 (\mod{4})</math>, <math>n</math> is odd and <math>>1</math> because <math>b>0</math>. Therefore, substituting for <math>x</math> and <math>y</math> we get: <math>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}</math>.
 +
 
 +
==Simplified Evan Chen's Solution==
 +
 
 +
Let <math>x = da</math>, <math>y = db</math>, such that <math>gcd(a,b) = 1</math>. It's obvious <math>x > y</math>, so <math>a > b</math>
 +
 
 +
By plugging it in to the original equation, and simplifying, we get
 +
 
 +
<cmath>d = \frac{ab(a^2 + 3b^2)(3a^2 + b^2)}{(a-b)^7}</cmath>
 +
 
 +
Since d is an integer, we got:
 +
 
 +
<cmath>(a-b)^7 \mid ab(a^2 + 3b^2)(3a^2 + b^2) \: \: \: \: \: (*)</cmath>
 +
 +
1. Since <math>gcd(a,b) = 1</math>, a and b can't both be even
 +
 
 +
2. If both a and b are odd.
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<math>a-b</math> will be even. The left hand side (*) has at least 7 factors of 2.
 +
Evaluating the (*) RHS term by term, we get:
 +
<cmath>a \equiv 1 \pmod{2}</cmath>
 +
<cmath>b \equiv 1 \pmod{2}</cmath>
 +
<cmath>a^2 + 3b^2 \equiv 4 \pmod{8}</cmath>
 +
<cmath>3a^2 + b^2 \equiv 4 \pmod{8}</cmath>
 +
 
 +
the (*) RHS has only 4 factors of 2. Contradiction.
 +
 
 +
3. Thus, <math>a-b</math> has to be odd. We want to prove  <math>a-b=1</math>
 +
 
 +
We know:
 +
<cmath>a \equiv b \pmod{a-b}</cmath>
 +
So
 +
<cmath>ab(a^2 + 3b^2)(3a^2 + b^2) \equiv  16a^6 \pmod{a-b}</cmath>
 +
We can assume if <math>a-b \neq 1</math>
 +
 
 +
Since <math>a-b</math> is odd,  <math>gcd(16, a-b) = 1</math>
 +
 
 +
Since <math>gcd(a, a-b) = 1</math>, <math>gcd(a^6, a-b) = 1</math>
 +
 
 +
<cmath>ab(a^2 + 3b^2)(3a^2 + b^2) \equiv  16a^6 \neq 0 \pmod{a-b}</cmath>
 +
contradiction.  So, <math>a-b = 1</math>.
 +
 
 +
If <math>a-b=1</math>, obviously (*) will work, d will be an integer.
 +
 
 +
So (*) <math>\Leftrightarrow</math>  <math>a-b=1</math>. The proof is complete.
 +
 
 +
-Alexlikemath
 +
 
 +
 
 
{{MAA Notice}}
 
{{MAA Notice}}
  
 
==See also==
 
==See also==
 
{{USAJMO newbox|year=2017|num-b=1|num-a=3}}
 
{{USAJMO newbox|year=2017|num-b=1|num-a=3}}

Revision as of 22:08, 4 April 2021

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. AMC logo.png

See also

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