TITLE: Show that $3x^4+4y^4=19z^4$ has no integer solution
QUESTION [7 upvotes]: If $x,y,z$ are integers such that $xyz\neq 0$, prove or disprove that
$$3x^4+4y^4=19z^4$$
has no solution.
Maybe this is can use quadratic residue to solve it, maybe this equation seem is famous? because I have solve paper Introduced this.

REPLY [3 votes]: $\color{brown}{\textbf{Version of 07.10.19.}}$
Is known that the Diophantine equation

$$uv=w^2,\quad (u,v,w)\in\mathbb N^3$$
  has the common solution
  $$u=k^2m,\quad v=l^2m,\quad w=klm,\quad \gcd(k,l)=1.$$

The issue equation can be presented in the form of
$$57(z^2-x^2)^2 + (8y^2)^2 = (3x^2-19z^2)^2,\tag1$$
or
$$57(z^2-x^2)^2 = (3x^2+8y^2-19z^2)(3x^2+8y^2-19z^2).\tag2$$
Equation
$$3x^2+8y^2\equiv 0\pmod{19}$$
has the single solution
$$x\equiv y \equiv 0\pmod{19}$$
(see also Wolfram Alpha solution),
and equation
$$8y^2-19z^2\equiv 0\pmod{3}$$
has the single solution
$$y\equiv z \equiv 0\pmod{3}$$
(see also Wolfram Alpha solution).
Besides, from $(1)$ should $z>x,$ so
$$
\begin{cases}
-3x^2+8y^2+19z^2 = m^2s\\
-3x^2-8y^2+19z^2 = 57n^2s\\
z^2-x^2=mns\\
(m,n,s)\in\mathbb N^2\\
\gcd(m,n)=1
\end{cases}\Rightarrow
\begin{cases}
16y^2 = (m^2+57n^2)s\\
2(19z^2-3x^2) = (m^2-57n^2)s\\
z^2-x^2=mns\\
(m,n,s)\in\mathbb N^2\\
\gcd(m,n)=1
\end{cases}
$$
\begin{cases}
16y^2 = (m^2+57n^2)s\\
32z^2 = (m^2-6mn-57n^2)s\\
32x^2 = (m^2-38mn-57n^2)s\\
(m,n,s)\in\mathbb N^2\\
\gcd(m,n)=1\tag3.
\end{cases}
From $(3.1),(3.2)$ unknowns $m,n$ should have the same parity.
Let $m=n+2p,$ then
\begin{cases}
8y^2 = (2p^2+2pn+29n^2)s\\
8z^2 = (p^2-2pn-13n^2)s\\
8x^2 = (p^2-18pn-13n^2)s\\
(n,p,s)\in\mathbb N^2\\
\gcd(n,2p)=1\tag4.
\end{cases}
From $(4.1),(4.5)$ should $8\,|\,s.$
Let $s=8t,$ then
\begin{cases}
y^2 = (2p^2+2pn+29n^2)t\\
z^2 = (p^2-2pn-13n^2)t\\
x^2 = (p^2-18pn-13n^2)t\\
(n,p,t)\in\mathbb N^2\\
\gcd(n,2p)=1\tag5.
\end{cases}
Since 
$$g=\gcd(p^2-2pn-13n^2,p^2-18pn-13n^2)\,|\,16pn,$$
then $g=1,$
$$\gcd(t,p^2-2pn-13n^2,p^2-18pn-13n^2)\,|\,16pn,$$
so $t=T^2,$
\begin{cases}
2p^2+2pn+29n^2=Y^2\\
p^2-2pn-13n^2 = Z^2\\
p^2-18pn-13n^2 = X^2\\
(n,p,t)\in\mathbb N^2\\
\gcd(n,2p)=1\tag6.
\end{cases}
Equation $(6.3),(6.5)$ can be written in the form of
$$Z^2 + 14n^2 = (p-1)^2,\quad 2\,|\not\ n.$$
By the parity, $Z=p-1-2r,$ and the system
$$2r^2-2(p-1)r+7n^2 =0,\quad 2\,|\not\ n$$
does not have solutions.

$$$$
Old version.
Is known that the Diophantine equation

$$kx^2+y^2=z^2,\quad (x,y,z,k)\in\mathbb N^4$$
  with odd $k,$ has common solution
  $$x=m(2n-1),\quad y=m\dfrac{(2n-1)^2-1}2,\quad z=m\dfrac{k(2n-1)^2+1}2,\quad (m,n)\in \mathbb N^2.$$

The issue equation can be presented in the form of
$$57(x^2-z^2)^2 + (8y^2)^2 = (3x^2-19z^2)^2,\tag1$$
with the common solution in the form of
$$\begin{cases}
x^2-z^2 = s^\,_1m(2n-1)\\
8y^2=m\dfrac{57(2n-1)2-1}2\\
3x^2-19z^2=s^\,_2m\dfrac{57(2n-1)^2+1}2\\
(m,n)\in\mathbb N^2,\quad (s^\,_1,s^\,_2)\in\{-1,1\}^2,
\end{cases}$$
or
$$\begin{cases}
x^2-z^2 = s^\,_1m(2n-1)\\
4y^2=m(57n^2-57n+14)\\
3x^2-19z^2=s^\,_2m(114n^2-114n+29)\\
(m,n)\in\mathbb N^2,\quad (s^\,_1,s^\,_2)\in\{-1,1\}^2.
\end{cases}\tag2$$
$\color{brown}{\mathbf{Cases\ "+ +","- +".}}$
$$
\begin{cases}
x^2-z^2 = s^\,_1m(2n-1)\\
4y^2=m(57n^2-57n+14)\\
3x^2-19z^2=m(114n^2-114n+29)\\
(m,n)\in\mathbb N^2
\end{cases}\Rightarrow
\begin{cases}
x^2-z^2 = s^\,_1m(2n-1)\\
3x^2-8y^2-19z^2 = m\\
3x^2+8y^2-19z^2 = 57m(2n-1)^2.
\end{cases}\tag3$$
From $(3)$ should
$$6x^2\equiv 3y^2\equiv m\pmod{19},$$
with the single solution
$$6x^2\equiv 3y^2\equiv m \equiv 0.\tag4$$
(see Wolfram Alpha tables)
Let $(x_0,y_0,z_0,m_0,n_0)$ is the solution with the least norm $N(x,y,z)=x^2+y^2+z^2.$
Then from $(4)$ should
$$x_0 = 19x_1,\quad y_0 = 19y_1,\quad 19\,|\, m_0,$$
and from $(3.1)$ should
$$z_0 = 19z_1, \quad a_0 = 361 m_1.$$
Easily to see, that $(x_1,y_1,z_1,m_1,n_0)$ is another solution, with the norm
$$N(x_1,y_1,z_1) = \frac1{361}N(x_0,y_0,z_0).$$
Therefore, solution with the least positive norm does not exist, and zero solution is unique.
$\color{brown}{\mathbf{Cases\ "+ -","- -".}}$
$$
\begin{cases}
x^2-z^2 = s^\,_1m(2n-1)\\
4y^2=m(57n^2-57n+14)\\
19z^2-3x^2=m(114n^2-114n+29)\\
(m,n)\in\mathbb N^2
\end{cases}\Rightarrow
\begin{cases}
x^2-z^2 = s^\,_1m(2n-1)\\
19z^2-3x^2-8y^2 = m\\
19z^2-3x^2+8y^2 = 57m(2n-1)^2.
\end{cases}\tag5$$
From $(5)$ should
$$13x^2\equiv16\equiv m\pmod{19},$$
with the single solution
$$6x^2\equiv 3y^2\equiv m \equiv 0\tag6$$
(see Wolfram Alpha table)
and the similar proof.
$\color{brown}{\mathbf{Proved.}}$