2

Prove that the equation $x^3+y^3+z^3= 2014^{2012}$ doesn't have integer solutions.

How does one come up with the idea of using congruences modulo $9$?

Batominovski
  • 49,629
meowy03
  • 415
  • 4
    Because $\phi(9)=2\cdot 3$, it follows that $t^3\equiv 0,\pm1 \pmod{9}$. If you see a power $k$ in a diophantine equation, you might want to find a small natural number $n$ s.t. $k\mid \phi(n)$ so that there are not so many residues of $t^k$ mod $n$. – Batominovski May 06 '20 at 12:48
  • @WETutorialSchool could you post this comment as an answer? – meowy03 May 08 '20 at 09:18

2 Answers2

3

Since $k=2014^{2012}\equiv 4\bmod 9$, we know that $k$ cannot be represented as the sum of three cubes - see The sum of three cubes.

Dietrich Burde
  • 130,978
1

Because $\phi(9)=2\cdot 3$, it follows that $t^3\equiv 0,\pm1 \pmod{9}$. If you see a power $k$ in a diophantine equation, you might want to find a small natural number $n$ s.t. $k\mid \phi(n)$ so that there are not so many residues of $t^k$ mod $n$. Usually, we only look for an integer $n$ which is a prime power.

Here are some related questions that can be solved by using the paragraph above.

  • Find all integers $x$, $y$, and $z$ s.t. $x^3+y^3=z^6+3$. (See also here.)

  • Find all integers $x$ and $y$ s.t. $x^5-3y^5=2008$. (See also here.)

  • Find all integers $x$ and $y$ s.t. $x^3+117y^3=5$. (See also here.)

  • Find all integers $x$ and $y$ s.t. $7x^3+2=y^3$. (See also here.)

  • Find all integers $x$ and $y$ s.t. $x^4-y^4=1996$. (See also here.)

Batominovski
  • 49,629