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>
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==Solution 2==
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Let <math>n = \frac{x+y}{3}</math>.
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Thus, <math>x+y = 3n</math>.
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We have
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<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>
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Substituting <math>n</math> for <math>\frac{x+y}{3}</math>, we have
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<cmath>9n^2 - x(3n-x) = (n+1)^3</cmath>
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Treating <math>x</math> as a variable and <math>n</math> as a constant, we have
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<cmath>9n^2 - 3nx + x^2 = (n+1)^3,</cmath>
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which turns into
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<cmath>x^2 - 3nx + (9n^2 - (n+1)^3) = 0,</cmath>
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a quadratic equation.
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By the quadratic formula,
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<cmath>x = \frac{1}{2} \left(3n \pm \sqrt{9n^2 - 4(9n^2 - (n+1)^3)} \right)</cmath>
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which simplifies to
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<cmath>x = \frac{1}{2} \left(3n \pm \sqrt{4(n+1)^3 - 27n^2} \right)</cmath>
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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.
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We can factor the aforementioned equation to be <cmath>(n-2)^2 (4n+1) = k^2</cmath>
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for an integer <math>k</math>.
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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.
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Since <math>4n + 1</math> is odd, the square root of the aforementioned equation must be odd as well.
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Thus, we have <math>4n + 1 = a^2</math> for some odd <math>a</math>.
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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.
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Thus, <cmath>x+y = 3n = 3p(p-1).</cmath>
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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>.
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(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.)
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Since <math>y = 3n - x</math>, the general form for <math>y</math> is just
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<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>
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(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.)
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for a positive integer <math>p</math>.
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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>
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<math>\boxed{}</math>
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-fidgetboss_4000

Latest revision as of 22:42, 15 March 2020

Problem

Solve in integers the equation \[x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3.\]

Solution

We first notice that both sides must be integers, so $\frac{x+y}{3}$ must be an integer.

We can therefore perform the substitution $x+y = 3t$ where $t$ is an integer.

Then:

$(3t)^2 - xy = (t+1)^3$

$9t^2 + x (x - 3t) = t^3 + 3t^2 + 3t + 1$

$4x^2 - 12xt + 9t^2 = 4t^3 - 15t^2 + 12t + 4$

$(2x - 3t)^2 = (t - 2)^2(4t + 1)$

$4t+1$ is therefore the square of an odd integer and can be replaced with $(2n+1)^2 = 4n^2 + 4n +1$

By substituting using $t = n^2 + n$ we get:

$(2x - 3n^2 - 3n)^2 = [(n^2 + n - 2)(2n+1)]^2$

$2x - 3n^2 - 3n = \pm (2n^3 + 3n^2 -3n -2)$

$x = n^3 + 3n^2 - 1$ or $x = -n^3 + 3n + 1$

Using substitution we get the solutions: $(n^3 + 3n^2 - 1, -n^3 + 3n + 1) \cup (-n^3 + 3n + 1, n^3 + 3n^2 - 1)$

Solution 2

Let $n = \frac{x+y}{3}$. Thus, $x+y = 3n$. We have \[x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3 \implies (x+y)^2 - xy = \left(\frac{x+y}{3}+1\right)^3\] Substituting $n$ for $\frac{x+y}{3}$, we have \[9n^2 - x(3n-x) = (n+1)^3\] Treating $x$ as a variable and $n$ as a constant, we have \[9n^2 - 3nx + x^2 = (n+1)^3,\] which turns into \[x^2 - 3nx + (9n^2 - (n+1)^3) = 0,\] a quadratic equation. By the quadratic formula, \[x = \frac{1}{2} \left(3n \pm \sqrt{9n^2 - 4(9n^2 - (n+1)^3)} \right)\] which simplifies to \[x = \frac{1}{2} \left(3n \pm \sqrt{4(n+1)^3 - 27n^2} \right)\] Since we want $x$ and $y$ to be integers, we need $4(n+1)^3 - 27n^2$ to be a perfect square. We can factor the aforementioned equation to be \[(n-2)^2 (4n+1) = k^2\] for an integer $k$. Since $(n-2)^2$ is always a perfect square, for $(n-2)^2 (4n+1)$ to be a perfect square, $4n + 1$ has to be a perfect square as well. Since $4n + 1$ is odd, the square root of the aforementioned equation must be odd as well. Thus, we have $4n + 1 = a^2$ for some odd $a$. Thus, \[n = \frac{a^2 - 1}{4},\] in which by difference of squares it is easy to see that all the possible values for $n$ are just $n = p(p-1)$, where $p$ is a positive integer. Thus, \[x+y = 3n = 3p(p-1).\] Thus, the general form for \[x = \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)\] for a positive integer $p$. (This is an integer since $4(p(p-1)+1)^3 - 27(p(p-1))^2$ is an even perfect square (since $4(p(p-1)+1)$ is always even, as well as $27(p(p-1))^2$ being always even) as established, and $3p(p-1)$ is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by $2$ always an integer.) Since $y = 3n - x$, the general form for $y$ is just \[y = 3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)\] (This is an integer since $4(p(p-1)+1)^3 - 27(p(p-1))^2$ is an even perfect square (since $4(p(p-1)+1)$ is always even, as well as $27(p(p-1))^2$ being always even) as established, and $3p(p-1)$ is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by $2$ always an integer, which thus trivially makes \[3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)\] an integer.) for a positive integer $p$. Thus, our general in integers $(x, y)$ is \[(\frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right), 3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right).\] $\boxed{}$

-fidgetboss_4000

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