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Difference between revisions of "2015 USAJMO Problems/Problem 2"

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(→‎Solution)
 
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Then:
 
Then:
  
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<math>(3t)^2 - xy = (t+1)^3</math>
+
<cmath>
 +
\begin{align*}
 +
(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)
 +
\end{align*}
 +
</cmath>
  
−
<math>9t^2 + x (x - 3t) = t^3 + 3t^2 + 3t + 1</math>
+
<math>4t+1</math> is therefore the square of an odd integer and can be replaced with <math>(2n+1)^2 = 4n^2 + 4n +1</math>.
−
 
 
−
<math>4x^2 - 12xt + 9t^2 = 4t^3 - 15t^2 + 12t + 4</math>
 
−
 
 
−
<math>(2x - 3t)^2 = (t - 2)^2(4t + 1)</math>
 
−
 
 
−
<math>4t+1</math> is therefore the square of an odd integer and can be replaced with <math>(2n+1)^2 = 4n^2 + 4n +1</math>
 
  
 
By substituting using <math>t = n^2 + n</math> we get:
 
By substituting using <math>t = n^2 + n</math> we get:
  
−
<math>(2x - 3n^2 - 3n)^2 = [(n^2 + n - 2)(2n+1)]^2</math>
+
<cmath>
−
 
+
\begin{align*}
−
<math>2x - 3n^2 - 3n = \pm (2n^3 + 3n^2 -3n -2)</math>
+
(2x - 3n^2 - 3n)^2 &= [(n^2 + n - 2)(2n+1)]^2 \\
−
 
+
2x - 3n^2 - 3n &= \pm (2n^3 + 3n^2 -3n -2) \\
−
<math>x = n^3 + 3n^2 - 1</math> or <math>x = - n^3 + 3n + 1</math>
+
x = n^3 + 3n^2 - 1 &\text{ or } x = - n^3 + 3n + 1
 +
\end{align*}
 +
</cmath>
  
 
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>

Latest revision as of 11:55, 21 February 2017

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:

\begin{align*} (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) \end{align*}

$4t+1$ 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:

\begin{align*} (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 &\text{ or } x = - n^3 + 3n + 1 \end{align*}

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

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