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 $5$.
 A: Look modulo 8.  Remember that odd squares are 1 more than a multiple of 8.
What are the possible remainders of $z^2-5x^2$?
What the the possible remainders of $2y(3x+y-z)$ if $y$ is odd? if $y$ is even?
A: 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 $(x,y,z)=(0,0,0)$.
A: $$ g(x,y,z) = 5x^2 + 2 y^2 - z^2 - 2 yz + 6 xy.  $$
$$ 2g = u^2 + v^2 - 12 w^2, $$ where
$$ u = 3x + 2 y - z; \; \;  v = x + 3 z; \; \;  w = z  $$ and the linear mapping that takes $(x,y,z) \mapsto (u,v,w)$ is nonsingular. 
There are no solutions (not all zero) with $u,v,w$ coprime.
Since $3$ divides $12,$ $3 | (u^2 + v^2),$ and $(-1|3) = -1$ tells us that $3|u,$ $3|v,$ and so $9|(u^2 + v^2).$ This then tells us that $3 |w.$ So, in fact, $3|u,v,w$ and they are not coprime. About $x,y,z,$ we get $3|z,$ then $3 |x,$ then $3 |y,$ all from the definitions of $u,v,w.$
Ummm. There are many complicated ways to describe "infinite descent" or, in this case, being anisotropic in $\mathbb Q_3.$ There is also a simple way: if there is a solution to $g=0$ with $x,y,z$ integers not all $0,$ then we may divide out by their gcd and arrive at a solution with $\gcd(x,y,z) = 1.$  We have proved there is no solution with $\gcd(x,y,z) = 1.$ That's it, the only integer solution is $(0,0,0)$
The global relation, pages 46 and 76 in CASSELS, says that an indefinite ternary quadratic form is anisotropic in an even number of $p$-adic fields
 in this case both $2,3.$ Actually, he does not specify the ternary case there, one needs to go back and apply using Lemma 2.5 on page 59.
AHA! This explains something from fifteen years ago. I mentioned to a couple of genuine quadratic forms experts that a positive ternary quadratic form is anisotropic for an odd number of (finite) primes, therefore at least one, and an indefinite form an even number of primes (possibly none). They expressed surprise. Looking again, this could be a homework question ( I do not think it is) in Cassels, and would not be among the more difficult exercises. But it is not stated explicitly in the main text, so I guess people don't know it. There we go, a somewhat abbreviated mention on page 142 in CONWAY. You can also purchase the real book, which I recommend.
