MAT102 Lecture 7

• 1
  • Show that $x^3+2y^3=4z^3$ has no positive integer solutions.
  • Let's proceed by contradiction.
  • a)
    • $x_0=2k$ for some $k\in \mathbb{Z}$
    • $(2k{)}^3+2y^3=4z^3$
    • $8k^3+2y^3=4z^3$
    • $8k^3-4z^3=-2y^3$
    • $4z^3-8k^3=2y^3$
    • $2(2z^3-4k^3=y^3)$
    • Now after isolating for $y$. We can see if y is either odd or even.
    • $2z^3-4k^3=y^3$
    • Since we can pull out a 2 from the left side, we know that the result of the left side will always be even.
    • $2(z^3-2k^3)=y^3$
    • This means that $y_0$ and $y^3$ is even.
    • Since $x_0$ is even, $y_0$ is even this means that $z_0$ must be even, since an even number + an even number = an even number based on previous course work.
  • b)
    • $x_0=2x_1$
    • $(2x_1{)}^3+(2y_0{)}^3=(4z_0{)}^3$
    • $4(x_1{)}^3+(y_0{)}^3=2(z_0{)}^3$
    • The same equation.
    • If you do the same work as part a), you can prove that either $y_0$ or $z_0$ is even.
  • c)
    • $x^3+2y^3=4z^3$
    • Let's show that there is always a solution smaller than $(x_0,y_0,z_0)$.
    • $(x_0{)}^3+2(y_0{)}^3=4(z_0{)}^3$
    • Let's set up some variables since we know every one of these is even.
    • $\{\begin{array}{l}{x}_0=2a\\ y_0=2b\\ z_0=2c\end{array}$
    • $(2a{)}^3+2(2b{)}^3=4(2c{)}^3$
    • $8a^3+16b^3=32c^3$
    • $8(a^3+2b^3)=8(4c^3)$
    • $a^3+2b^3=4c^3$
    • If
      • $x_0=2a$
      • $(x_0{)}^3=(2a{)}^3$
      • $(x_0{)}^3=8a^3$
      • However after simplifying we have $a^3$, which means that $a^3<(x_0{)}^3$, because $(x_0{)}^3=8a^3$.
    • We can do this again.
      • We know that $a,b,c$ are all even. Meaning if we do:
      • $\{\begin{array}{l}a=2l\\ b=2m\\ c=3n\end{array}$
      • Resulting in an endless cycle.
    • If we can always find a smaller solution, that means that there is no solution.
    • This is the same logic we used to figure out that there is no smallest real number.
    • If there is no smallest positive integer solution, then there is none.
    • Contradicts the well-ordering principle that there is a smallest element.
      • Our smallest element should be 1 as an integer since we have a limited set of numbers to choose, (positive integer solutions).

    ---

    • Define $X=\{x_i,i\in \mathbb{N}\}\subseteq \mathbb{N}$ and it's non-empty
    • By the well ordering principle it has to have a minimal element like $x_n$.
    • $x_{n+1}<x_n$, which is a contradiction.