Can we show the failure of Hasse principle (or "local-global" principle) for these simultaneous quadratic equations?
\begin{eqnarray} x^2 - 5y^2 &=& uv \tag{1}\\ x^2 - 5z^2 &=& (u+v)(u+2v) \tag{2} \end{eqnarray} These are homogeneous polynomials so the left side variables are $[x:y:z] \in \mathbb{P}^2$ while the right side variables are $[u:v] \in \mathbb{P}^1$.
Two quadratics in 5 variables; this is a del Pezzo surface of degree $4$.
- two-dimensional Fano variety; non-singular projective algebraic surface with ample anti-canonical divisor class (Here I just looked up the word "divisor" from algebraic geometry to see what it means here).
- intersection of two conics (in this case) $X \subseteq \mathbb{P}^4$.
"Failure" here means that $X(\mathbb{Q})= \varnothing$ is empty even though $X(\mathbb{Q}_p)$ has points for all primes $p \in \mathbb{Z}$. The lack of solutions is parametrized by the Brauer group. In fact,$$\text{Br}(X)/\text{Br}(\mathbb{Q}) \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$ So we can ask for adelic solutions that don't converge to rational solutions.
The Brauer group of a field would be defined by the exact sequence:
$$ 0 \to \text{Br}(K) \to \bigoplus_{v \in S} \text{Br}(K_v) \to \mathbb{Q}/\mathbb{Z} \to 0 $$
The group splits by some sort of prime factorization and it's value is a rational number. This is related to Class Field Theory. It's also called a "Galois cohomology" like this: $$ \text{Br}(K) \simeq H^2(K, \mathbf{G}_m) $$ where $\mathbf{G}_m$ is the "multiplicative group", $\times$.