show this diophantine equation $$3x^4-y^2=3$$ has no integer $(y\neq 0)$ solution?
My try: WLOG Assmue $(x,y)$ is postive integer solution,then $3|y$,let $y=3y'$,then we have $$x^4-1=3y'^2\tag{1}$$ and following I want $\pmod5$,since $x^4\equiv 0,1\pmod 5$
(1):if $x^4\equiv 0\pmod 5$or $5|x$,then $(1)$ it is clear no solution
(2)if $(x,5)=1$,then $x^4-1\equiv 0\pmod 5$,there also exsit $y'$ such $5|3y'^2$,so How to solve my problem? Thanks
I tried the following.
$3x^4-y^2=3$
$3x^4-3=y^2$
$3(x^2-1)(x^2+1)=y^2 \tag{1}$
If $y^2$ is divisible by $3$ then one of the factors of $(x^2-1)(x^2+1)$ should be divided by $3$ to make the right hand side a perfect square. So $x$ can't be $0 \pmod{3}$. Now if we divide 3 from the both hand side we can see the following.
$$ (x^2-1)(x^2+1)=3^{\alpha}\cdot B^{\gamma} \tag{2} $$
Note that $\alpha$ is odd and $B$ is not necessarily a prime factor.
We've already argued $x\not\equiv 0\pmod{3}$, so $x\equiv +1\pmod{3}$ or $x\equiv -1\pmod{3}$.
If $x\equiv +1\pmod{3}$ then $x^2-1=3^{2\lambda}\cdot D^\eta$ and $x^2+1\not\equiv 0\pmod{3}$. So $\alpha\ne2\lambda$.
If $x\equiv -1 \pmod{3}$, then $x^2+1=3^{2\lambda}\cdot D^\eta$ and $x^2-1\not\equiv 0\pmod{3}$. So $\alpha\ne2\lambda$.
Hope this helps.
$(1)$ WLOG we study the impossibility of $A^4-3B^2=1$ with $B\ne0$. Clearly $(A,B)=1$ and by modulo $5$ and modulo $16$ we have $A$ odd and $10|B$.
$(2)$ We need only consider non-negative solutions. The fundamental unit of the ring $\mathbb Z[\sqrt3]$ is $2+\sqrt3$ so all (positive) solution $(x_n,y_n)$ of the equation $X^2-3Y^2=1$ is given by $x_n+y_n\sqrt3=(2+\sqrt3)^n$.
$(3)$ Looking modulo $10$ at the $(x_n,y_n)$ we find that there is a periodicity of order 6. In fact we have $$\begin{cases}x_1+y_1\sqrt3=2+\sqrt3\\x_2+y_2\sqrt3=7+4\sqrt3\\x_3+y_3\sqrt3=26+15\sqrt3\\x_4+y_4\sqrt3=97+56\sqrt3\\x_5+y_5\sqrt3=362+209\sqrt3\\x_6+y_6\sqrt3=1351+780\sqrt3\end{cases}\Rightarrow\begin{cases}(x_1,y_1)\equiv(2,1)\\(x_2,y_2)\equiv(7,4)\\(x_3,y_3)\equiv(6,5)\\(x_4,y_4)\equiv(7,6)\\(x_5,y_5)\equiv(2,9)\\(x_6,y_6)\equiv(1,0)\end{cases}$$ from which we have $$(x_{6k+h},y_{6k+h})\equiv(x_h,y_h)\pmod{10}$$
$(4)$ Consequently since $A^4-3B^2=1$ and $B\equiv0\pmod{10}$ we must have $$A^2+B\sqrt3=(2+\sqrt3)^{6n}$$ Consider $a_{2n}+b_{2n}\sqrt3=(2+\sqrt3)^{2n}$ so we have $$A^2+B\sqrt3=(a_{2n}+b_{2n}\sqrt3)^3\Rightarrow A^2=a_{2n}^3+9a_{2n}b_{2n}^2=a_{2n}(a_{2n}^2+(3b_{2n})^2)$$.
$(5)$ Since $a_{2n}^2-3b_{2n}^2=1$ one has $(a_{2n},b_{2n})=1$ and because $3$ can not divide $a_{2n}$ we deduce the factors $a_{2n}$ and $a_{2n}^2+(3b_{2n})^2$ of $A^2$ are coprime therefore $a_{2n}=\alpha_{2n}^2$ for some positive integer $\alpha$ (less than $a_{2n}$ which is distinct of $1$ because the smallest possible positive value of $a_{2n}$ is $2$) whereby we get $$a_{2n}^2-3b_{2n}^2=1\Rightarrow \alpha_{2n}^4-3b_{2n}^2=1$$
Given this last equation, it is clearly seen that we can apply Fermat's method of infinite descent. We are done!
This curve $y^2=3x^4-3$ has only two integral points, according to the magma online calculator as follows.
IntegralQuarticPoints($[3,0,0,0,-3]$);
It says that all integral points are $( \pm 1 , 0)$.
Hence this equation has no integer $(y\neq 0)$ solution.
See related info MO
