Can someone help with a task? Need to find a solution other than $(0,0,0) $ with infinite descent. $x,y,z\in\mathbb{Z}$. Any help would be appreciated.
The equation is $x^2-3y^2=2z^2$. I tried to write it down as $x^2 ≡y^2 ≡0 \pmod 2$. Then $x^2 =a\cdot2$ and $y^2=b\cdot2$. Then $a\cdot2-3\cdot b\cdot 2=2z^2.$ And it doesn't work any further
First, consider what the argument of infinite descent signifies.
The set of all positive integers are bounded below, by the number $(+1)$, which is the smallest possible positive integer.
Suppose that it could be shown that (1) above has a solution in positive integers. You could let the set $S$ denote the set of all ordered triples $\{(x_1,y_1,z_1), (x_2,y_2,z_2), \cdots \}$ that represent a positive integer solution to (1) above. Then, from this set $S$, you could order the elements by the size of the first component of each ordered triple.
That is, you could construe that the solution $(x_1, y_1, z_1) \leq $ the solution $(x_2, y_2, z_2)$ if and only if $x_1 \leq x_2$.
Then, because the positive integers are bounded below, there would have to be some specific value $X$ which represents the minimum value of the first component of any of the ordered triples in the set $S$.
Suppose further, that it could be shown that for any element $(X,Y,Z)$ that is in the set $S$, that $~\displaystyle \left(\frac{X}{3}, \frac{Y}{3}, \frac{Z}{3}\right)~$ is also in the set $S$.
This would contradict the assertion that the set $S$ has a smallest element. This situation would therefore be impossible, because the positive integers are bounded below by $(+1)$.
Therefore, it is sufficient to show that if $(X,Y,Z)$ is an element in the set $S$, that so is $~\displaystyle \left(\frac{X}{3}, \frac{Y}{3}, \frac{Z}{3}\right)$.
So, the problem reduces to showing that
$$(X,Y,Z) \in S \implies \left(\frac{X}{3}, \frac{Y}{3}, \frac{Z}{3}\right) \in S. \tag2 $$
To show, (2) above, first I will establish some preliminary results. Then, I will use these preliminary results.
Preliminary Results
$\underline{\text{Lemma 1}: ~n \in \Bbb{Z^+} \implies n^2 \not\equiv 2\pmod{3}}$
Proof
When any positive integer $n$ is divided by $3$, the remainder will have to be some element $r$ in $\{0,1,2\}$. Therefore, Lemma 1 may be established by verifying that it holds for all three possible values for $r$.
$(3k + 0)^2 = 9k^2 + 0k + 0^2 \equiv 0 \pmod{3}.$
$(3k + 1)^2 = 9k^2 + 6k + 1^2 \equiv 1 \pmod{3}.$
$(3k + 2)^2 = 9k^2 + 12k + 2^2 \equiv 4 \pmod{3} \equiv 1 \pmod{3}.$
$\underline{\text{Lemma 2}: ~n \in \Bbb{Z^+}, 2n \equiv 1\pmod{3} \implies n \equiv 2\pmod{3}}$
Proof
Use the same type of approach as was used in Lemma 1.
$2 \times (3k + 0) = 6k \equiv 0 \pmod{3}.$
$2 \times (3k + 1) = 6k + 2 \equiv 2 \pmod{3}.$
$2 \times (3k + 2) = 6k + 4 \equiv 4 \pmod{3} \equiv 1 \pmod{3}.$
Only the last case, $(3k + 2)$ generated a product that was congruent to $(1) \pmod{3}.$
$\underline{\text{Lemma 3}: ~n \in \Bbb{Z^+}, n \equiv 0\pmod{3} \implies n^2 \equiv 0\pmod{9}}$
Proof
$n \equiv 0\pmod{3}$ implies that $n$ may be expressed as $(3k)$.
Then $n^2 = 9k^2$ which is a multiple of $(9)$.
Completion of the Problem
Let $(x,y,z)$ be any element in $S$.
Then, either $x$ is a multiple of $3$, or it is not a multiple of $3$.
Suppose that $x$ is not a multiple of $3$.
Then, by Lemma 1, $x^2 \equiv 1\pmod{3}.$
So, by assumption, you have that :
This implies that $2z^2 \equiv 1\pmod{3}.$
By Lemma 2, this implies that $z^2 \equiv 2\pmod{3}.$
By Lemma 1, this is impossible.
Therefore, for any element $(x,y,z) \in S$, the assumption that $x$ is not a multiple of $3$ has led to a contradiction.
Therefore, $x$ must be a multiple of $3$.
So, this implies that $(x^2 - 3y^2)$ is also a multiple of $3$.
Therefore, $2z^2$ must be a multiple of $(3)$.
Therefore. $z^2$ must be a multiple of $(3)$.
Therefore, $z$ must be a multiple of $(3)$.
So, it has been established that if $(x,y,z)$ is any element in $S$, that both $x$ and $z$ must be multiples of $3$.
By Lemma 3, this implies that both $x^2$ and $z^2$ are multiples of $9$.
Therefore, so is $2z^2$.
Also, you have that $x^2 - 3y^2 = 2z^2$.
Therefore, $3y^2$ must be a multiple of $9$.
Therefore $y^2$ must be a multiple of $3$.
Therefore, $y$ must be a multiple of $3$.
So, it has been established that if $(x,y,z)$ is any element in $S$, that each of $x,y,z$ must be a multiple of $3$.
This implies that the ordered triplet $~\displaystyle \left(\frac{x}{3}, \frac{y}{3}, \frac{z}{3}\right)~$ is an ordered triplet of positive integers. This begs the question: is this ordered triplet an element in $S$.
Since $(x,y,z)$ is an element in $S$, you know that
$$x^2 - 3y^2 = 2z^2.$$
This implies that
$$\frac{x^2}{9} - 3\left(\frac{y^2}{9}\right) = 2 \left(\frac{z^2}{9}\right).$$
This implies that
$$\left(\frac{x}{3}\right)^2 - 3\left(\frac{y}{3}\right)^2 = 2 \left(\frac{z}{3}\right)^2.$$
This implies that $~\displaystyle \left(\frac{x}{3}, \frac{y}{3}, \frac{z}{3}\right)~$ is a positive integer solution that is in $S$.
Therefore, you have that given any element $(x,y,z)$ in the set $S$, that the ordered triplet $~\displaystyle \left(\frac{x}{3}, \frac{y}{3}, \frac{z}{3}\right)~$ is also in the set $S$.
Consequently the assertion expressed in (2) above has been established.
Consequently, it has been established, that the equation in (1) above can not have any solution in positive integers.