Show that $3x^4+4y^4=19z^4$ has no integer solution

Question

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.

Answer 1

The equation is $$ 3x^4+4y^4=19z^4,\tag 1 $$ where $x,y,z$, integers. We also assume that $$ xyz\neq 0\textrm{ and }(x,y,z)=1.\tag 2 $$

We write (1) in the form $$ 3x^4-3z^4=16z^4-4y^4. $$ Hence $$ 3(x^2-z^2)(x^2+z^2)=4(2z^2-y^2)(2z^2+y^2) $$ or equivalently $$ N=3(x-z)(x+z)(x^2+z^2)=4(2z^2-y^2)(2z^2+y^2). $$ Obviously $N\neq 0$ from (1) and (2). If $xyz\neq0$, then $x^2+z^2>1$ and $2z^2+y^2>1$ and $4|N$, $3|N$.

Also $x,z$ not both even (since $(x,y,z)=1$) and writing $x=2k+1$, $z=2l$, we get ($x^4=16k_1+1$): $$ N=3(x^4-z^4)=3(16k_1+1-16l^4)=4(8l^2-y^2)(8l^2+y^2), $$ which is imposible. The other case is if $x=2k$ and $z=2l+1$, then ($z^2=4l_1+1,z^4=16l_2+1$): $$ N=3(x^4-z^4)=3(16k^4-16l_2-1)=4(8l_2+2-y^2)(8l_2+2+y^2), $$ which is also imposible.

If $x,z$ both odd, then $x=2k+1$, $z=2l+1$, $8|\left(4(2z^2-y^2)(2z^2+y^2)\right)\Rightarrow 2|y$. Hence $y=2m$ and $x^2=8k_1+1$,$z^2=8l_1+1$, $x^4=16k_2+1,z^4=16l_2+1$. Hence $3(x^4-z^4)=3\cdot 16(k_2-l_2)=4(2(8l_1+1)-4m^2)(2(8l_1+1)+4m^2)\Rightarrow$ $3\cdot4(k_2-l_2)=(8l_1+2-4m^2)(8l_1+2+4m^2)\Rightarrow$ $3(k_2-l_2)=(4l_1+1-2m^2)(4l_1+1+2m^2)=(4l_1+1)^2-4m^4$.

But $k_2=4k_1^2+k_1$, $l_2=4l_1^2+l_1$. Hence $(k_2-l_2)=4(k_1^2-l_1^2)+(k_1-l_1)$. If $k_1-l_1=1(mod4)$, then contradiction (easily). If $k_1-l_1=3(mod4)$, then I have make error in the proof and the problem is open. I hope in the next few days I find a proof.

Answer 2

$\color{brown}{\textbf{Version of 07.10.19.}}$

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.

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$$$$ 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.

$\color{brown}{\mathbf{Proved.}}$

Answer 3

COMMENT.- Just for fun. Do you have for the equation $3x^4+7y^4=19z^4$ the following irrational parameterization $$\begin{cases}x=\left(\sqrt[4]{\dfrac{19}{3}}\sqrt{\dfrac{1-t^2}{1+t^2}}\right)z\\y=\left(\sqrt[4]{\dfrac{19}{7}}\sqrt{\dfrac{2t}{1+t^2}}\right)z\end{cases}$$ There is a value of $t$ such that both $\sqrt[4]{\dfrac{19}{3}}\sqrt{\dfrac{1-t^2}{1+t^2}}$ and $\sqrt[4]{\dfrac{19}{7}}\sqrt{\dfrac{2t}{1+t^2}}$ be rational $\ne0$?

If the equation $3x^4+7y^4=19z^4$ is not solvable, certainly NOT. But if the equation is solvable, certainly YES. For example for the equation $10x^4+6y^4=285z^4$ we have the parameterization $$\begin{cases}x=\left(\sqrt[4]{\dfrac{285}{10}}\sqrt{\dfrac{1-t^2}{1+t^2}}\right)z\\y=\left(\sqrt[4]{\dfrac{285}{6}}\sqrt{\dfrac{2t}{1+t^2}}\right)z\end{cases}$$ and for $t=\pm\dfrac15\sqrt{\dfrac{179\pm12\sqrt{114}}{5}}$ (which is root of $125t^4-358t^2+125=0$) we get the solution $(x,y,z)=(3,5,2)$.