<s>[INST] Why does $5x^2+6xy+2y^2=2yz+z^2$ have no integer solutions? Why does $5x^2+6xy+2y^2=2yz+z^2$ have no primitive integer solutions?
Modulo $2$ says that $x$ and $z$ are odd.
Modulo $3$ says that $x=0 \bmod 3$ and $y=-z \bmod 3$.
I cannot get anything modulo $5$.
 [/INST] Clearly $x\equiv z\pmod{2}$.
If $x,z$ are both even, then $y$ must be even too (by checking $\pmod{4}$).  
By dividing both sides by $4$ we can reduce all $x,y,z$ by the prime $2$, unless at least one of $x,y,z$ is $0$.  
If $x=0$, then $2y^2=2yz+z^2$. Let $z=2k, k\in\mathbb Z$ (since $z$ is even). Then $y^2-2ky-2k^2=0$, but $\Delta=12k^2$, which is not a perfect square unless $k=0\iff z=0$ and thus $(x,y,z)=(0,0,0)$. 
If $y=0$, then $5x^2=z^2\iff x=z=0$.  
If $z=0$, then $5x^2+6xy+2y^2=0$. But $\Delta=-4y^2<0$, unless $y=0$, giving $(x,y,z)=(0,0,0)$.
So assume WLOG that $x,z$ are both odd. 
Then, since $x^2\equiv z^2\equiv 1\pmod{4}$, we have  
$$2xy+2y^2\equiv 2yz\pmod{4}\iff y(x+y)\equiv yz\pmod{2}$$
$$\iff y(1+y)\equiv y\pmod{2}\iff y+y^2\equiv y\pmod{2}\iff y^2\equiv 0\pmod{2}$$
So $y$ is even. Let $y=2m, m\in\mathbb Z$.
$$5x^2+12xm+8m^2=4zm+z^2$$  
Since $x^2\equiv z^2\equiv 1\pmod{8}$, we have
$$4+4xm\equiv 4zm\pmod{8}\iff 1+xm\equiv zm\pmod{2}\iff 1+m\equiv m\pmod{2}\iff 1\equiv 0\pmod{2}$$
Thus the only solution is $(x,y,z)=(0,0,0)$.
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