# Difference between revisions of "2015 USAMO Problems/Problem 1"

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Using substitution we get the solutions: <math>(n^3 + 3n^2 - 1, -n^3 + 3n + 1) \cup (-n^3 + 3n + 1, n^3 + 3n^2 - 1)</math> | Using substitution we get the solutions: <math>(n^3 + 3n^2 - 1, -n^3 + 3n + 1) \cup (-n^3 + 3n + 1, n^3 + 3n^2 - 1)</math> | ||

+ | |||

+ | ==Solution 2== | ||

+ | Let <math>n = \frac{x+y}{3}</math>. | ||

+ | Thus, <math>x+y = 3n</math>. | ||

+ | We have | ||

+ | <cmath>x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3 \implies (x+y)^2 - xy = \left(\frac{x+y}{3}+1\right)^3</cmath> | ||

+ | Substituting <math>n</math> for <math>\frac{x+y}{3}</math>, we have | ||

+ | <cmath>9n^2 - x(3n-x) = (n+1)^3</cmath> | ||

+ | Treating <math>x</math> as a variable and <math>n</math> as a constant, we have | ||

+ | <cmath>9n^2 - 3nx + x^2 = (n+1)^3,</cmath> | ||

+ | which turns into | ||

+ | <cmath>x^2 - 3nx + (9n^2 - (n+1)^3) = 0,</cmath> | ||

+ | a quadratic equation. | ||

+ | By the quadratic formula, | ||

+ | <cmath>x = \frac{1}{2} \left(3n \pm \sqrt{9n^2 - 4(9n^2 - (n+1)^3)} \right)</cmath> | ||

+ | which simplifies to | ||

+ | <cmath>x = \frac{1}{2} \left(3n \pm \sqrt{4(n+1)^3 - 27n^2} \right)</cmath> | ||

+ | Since we want <math>x</math> and <math>y</math> to be integers, we need <math>4(n+1)^3 - 27n^2</math> to be a perfect square. | ||

+ | We can factor the aforementioned equation to be <cmath>(n-2)^2 (4n+1) = k^2</cmath> | ||

+ | for an integer <math>k</math>. | ||

+ | Since <math>(n-2)^2</math> is always a perfect square, for <math>(n-2)^2 (4n+1)</math> to be a perfect square, <math>4n + 1</math> has to be a perfect square as well. | ||

+ | Since <math>4n + 1</math> is odd, the square root of the aforementioned equation must be odd as well. | ||

+ | Thus, we have <math>4n + 1 = a^2</math> for some odd <math>a</math>. | ||

+ | Thus, <cmath>n = \frac{a^2 - 1}{4}, </cmath> in which by difference of squares it is easy to see that all the possible values for <math>n</math> are just <math>n = p(p-1)</math>, where <math>p</math> is a positive integer. | ||

+ | Thus, <cmath>x+y = 3n = 3p(p-1).</cmath> | ||

+ | Thus, the general form for <cmath>x = \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)</cmath> for a positive integer <math>p</math>. | ||

+ | (This is an integer since <math>4(p(p-1)+1)^3 - 27(p(p-1))^2</math> is an even perfect square (since <math>4(p(p-1)+1)</math> is always even, as well as <math>27(p(p-1))^2</math> being always even) as established, and <math>3p(p-1)</math> is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by <math>2</math> always an integer.) | ||

+ | Since <math>y = 3n - x</math>, the general form for <math>y</math> is just | ||

+ | <cmath>y = 3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)</cmath> | ||

+ | (This is an integer since <math>4(p(p-1)+1)^3 - 27(p(p-1))^2</math> is an even perfect square (since <math>4(p(p-1)+1)</math> is always even, as well as <math>27(p(p-1))^2</math> being always even) as established, and <math>3p(p-1)</math> is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by <math>2</math> always an integer, which thus trivially makes <cmath>3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)</cmath> an integer.) | ||

+ | for a positive integer <math>p</math>. | ||

+ | Thus, our general in integers <math>(x, y)</math> is <cmath>(\frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right), 3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right).</cmath> | ||

+ | <math>\boxed{}</math> | ||

+ | |||

+ | -fidgetboss_4000 |

## Latest revision as of 22:42, 15 March 2020

### Problem

Solve in integers the equation

### Solution

We first notice that both sides must be integers, so must be an integer.

We can therefore perform the substitution where is an integer.

Then:

is therefore the square of an odd integer and can be replaced with

By substituting using we get:

or

Using substitution we get the solutions:

## Solution 2

Let . Thus, . We have Substituting for , we have Treating as a variable and as a constant, we have which turns into a quadratic equation. By the quadratic formula, which simplifies to Since we want and to be integers, we need to be a perfect square. We can factor the aforementioned equation to be for an integer . Since is always a perfect square, for to be a perfect square, has to be a perfect square as well. Since is odd, the square root of the aforementioned equation must be odd as well. Thus, we have for some odd . Thus, in which by difference of squares it is easy to see that all the possible values for are just , where is a positive integer. Thus, Thus, the general form for for a positive integer . (This is an integer since is an even perfect square (since is always even, as well as being always even) as established, and is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by always an integer.) Since , the general form for is just (This is an integer since is an even perfect square (since is always even, as well as being always even) as established, and is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by always an integer, which thus trivially makes an integer.) for a positive integer . Thus, our general in integers is

-fidgetboss_4000