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 | + | ==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 | + | 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\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>. |
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==Solution 2 (and motivation)== | ==Solution 2 (and motivation)== |
Revision as of 11:15, 30 January 2018
Problem:
Consider the equation
(a) Prove that there are infinitely many pairs of positive integers satisfying the equation.
(b) Describe all pairs of positive integers satisfying the equation.
Solution 1
We have , which can be expressed as . At this point, we think of substitution. A substitution of form is slightly derailed by the leftover x and y terms, so instead, seeing the xy in front, we substitute . This leaves us with , so . Expanding yields . Rearranging, we have . To satisfy this equation in integers, must obviously be a 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 does not work. We go to the next highest odd power, or . In this case, , so the LHS is , so . Using the original substitution yields as the first solution. We have shown part a by showing that there are an infinite number of positive integer solutions for , which can then be manipulated into solutions for . To solve part b, we look back at the original method of generating solutions. Define and to be the pair representing the nth solution. Since the nth odd number is , . It follows that . From our original substitution, .
Solution 2 (and motivation)
First, we shall prove a lemma:
LEMMA:
PROOF: Expanding and simplifying the right side, we find that which proves our lemma.
Now, we have that
Rearranging and getting rid of the denominator, we have that
Factoring, we have Dividing both sides, we have
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 . By inspection, must be a positive odd integer satistisfying . We also have Now, we can solve for and in terms of :
and .
Now we have:
and it is trivial to check that this parameterization works for all such (to keep and integral), which implies part (a).
MOTIVATION FOR LEMMA:
I expanded the LHS, noticed the coefficients were , 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.
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 |