The system of equations: $\begin{cases} 2x^2=4y^2+3z^2+2; \\ 13x=4y+3z+29 \end{cases}$

Question

Solve in positive integers the system of equations: $$\begin{cases} 2x^2=4y^2+3z^2+2; \\ 13x=4y+3z+29 \end{cases}$$

My work so far:

I used wolframalpha: $x=3,y=1,z=2$.

Answer 1

$$2x^2=4y^2+3z^2+2\tag1$$ $$13x=4y+3z+29\tag2$$

From $(1)(2)$, $$13^2(4y^2+3z^2+2)=2(13x)^2=2(4y+3z+29)^2,$$ i.e. $$644y^2+(-464-48z)y+489z^2-348z-1344=0$$ See this as a quadratic equation on $y$.

Since the discriminant has to be larger than or equal to $0$, we have to have $$(-464-48z)^2-4\cdot 644(489z^2-348z-1344)\ge 0,$$ i.e. $$\small -1.5\approx -\frac{236}{155}=\frac{2(87-49\sqrt{81})}{465}\le \frac{2(87-49\sqrt{69})}{465}\le z\le \frac{2(87+49\sqrt{69})}{465}\le \frac{2(87+49\sqrt{81})}{465}=\frac{352}{155}\approx 2.3$$ giving $$z=1,2$$

I think that you can continue from here.

Answer 2

$$2x^2=4y^2+3z^2+2\tag1$$ $$13x=4y+3z+29\tag2$$ (1) $$2x^2=4y^2+3z^2+2 \Rightarrow x>\max\{y,z\}$$ (2) $$13x=4y+3z+29<4x+3x+29 \Rightarrow 6x<29 \Rightarrow x<5$$

(1) $z -$ even. (2) $x -$ odd. Hence $x \in \{1,3\}$. But $x>y,z\ge1$. Hence $$x=3$$