Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2008 iTest Problems/Problem 91

Problem

Find the sum of all positive integers $n$ such that $x^3+y^3+z^3=nx^2y^2z^2$ is satisfied by at least one ordered triplet of positive integers $(x,y,z)$.

Solution (credit to official solution)

WLOG, let $x \ge y \ge z$. We know that $x^3 + x^3 + x^3 \ge x^3 + y^3 + z^3$, so \begin{align*} 3x^3 &\ge nx^2 y^2 z^2 \\ x &\ge n \cdot \frac{y^2 z^2}{3} \end{align*} We also know that $x^3 + y^3 + z^3 \equiv 0 \pmod{x^2}$, so $y^3 + z^3 \equiv 0 \pmod{x^2}$. Since $x,y,z$ are positive, $y^3 + z^3 \ge x^2$. Thus, \begin{align*} 2y^3 &\ge y^3 + z^3 \ge x^2 \ge n^2 \cdot \frac{y^4 z^4}{9} \\ 2 &\ge n^2 \cdot \frac{y z^4}{9} \\ 18 &\ge n^2yz^4 \end{align*} Note that if $z > 1$, then $n^2yz^4 \ge 32$, so $z = 1$. Thus, $y \le \frac{18}{n^2}$, making $y \le 4$. Now perform casework on the values of $y$.

  • If $y = 1$, then $2 \equiv 0 \pmod{x^2}$. Thus, $2$ must be a multiple of $x^2$. The only value of $x$ that works is $x = 1$, so $n = \tfrac{1+1+1}{1} = 3$.
  • If $y = 2$, then $9 \equiv 0 \pmod{x^2}$. Thus, $9$ must be a multiple of $x^2$. The only value of $x$ greater than or equal to $2$ that works is $x = 3$, so $n = \tfrac{1+2+3}{6} = 1$.
  • If $y = 3$, then $28 \equiv 0 \pmod{x^2}$. Thus, $28$ must be a multiple of $x^2$. However, there are no values of $x$ greater than or equal to $3$ that work.
  • If $y = 4$, then $65 \equiv 0 \pmod{x^2}$. Thus, $65$ must be a multiple of $x^2$. However, there are no values of $x$ greater than or equal to $4$ that work.

The only values of $n$ that results in the equation solvable by positive integers is $n = 1$ and $n = 3$, so the answer is $1 + 3 = \boxed{4}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 90 		Followed by:
Problem 92
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_91&oldid=99903"