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 (and motivation)== |
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 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>. | ||
{plshalp} | {plshalp} |
Revision as of 16:28, 17 June 2017
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 (and motivation)
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, . {plshalp} 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 |