= Solution
For massless up and down <quarks>, the <Quantum chromodynamics> kinetic term splits into two independent chirality sectors,
$$
\mathcal L_q=i\bar q_L\gamma^\mu D_\mu q_L+i\bar q_R\gamma^\mu D_\mu q_R.
$$
The colour <covariant derivative> acts identically on both flavours and commutes with constant flavour matrices. Independent transformations $q_R\mapsto Aq_R$, $q_L\mapsto Bq_L$, with $A,B\in SU(2)$, therefore leave the action invariant. This is <chiral symmetry>. The non-singlet axial transformations have traceless flavour generators, so they are not broken by the colour axial anomaly; the anomalous axial $U(1)$ is not being included in this symmetry.
For $V_{ij}=\bar q_{Lj}q_{Ri}$, transforming each index gives
$$
\boxed{V\mapsto AVB^\dagger=AVB^{-1}.}
$$
The nonzero condensate $\langle V\rangle=-vI$ is unchanged precisely when $A=B$. Thus <chiral symmetry breaking> leaves the diagonal vector subgroup $SU(2)_V$. Three of the six generators are broken, producing three massless <Goldstone bosons> in the exact massless limit. They are the <pions>, represented at low energy by the coset coordinate $U\in SU(2)$.
The <uniqueness of the two-derivative two-flavour chiral action> follows from the symmetry, rather than merely from guessing an invariant. The chiral group acts transitively on $U$, so every derivative-free invariant is constant. At $U=I$, the stabilizer acts on the three tangent directions by conjugation, the irreducible rotation representation. Its invariant symmetric quadratic form is a single multiple of $\delta_{ab}$. Transporting that form around the group gives a unique invariant two-derivative metric. Equivalently, the <Maurer-Cartan form> $U^\dagger\partial_\mu U$ is traceless, so an independent double-trace term vanishes; the remaining contracted single trace is the same metric. Lorentz invariance supplies the spacetime contraction. Up to total derivatives, a constant term and its overall coefficient, the massless action is therefore
$$
\boxed{\mathcal L_{\pi}=\frac{F^2}{4}\operatorname{tr}(\partial^\mu U^\dagger\partial_\mu U).}
$$
The positive coefficient is conveniently expressed through the <pion decay constant> $F$.
Use the <Pauli matrices> with $\operatorname{tr}(\tau_a\tau_b)=2\delta_{ab}$. Expanding the <matrix exponential> gives $\partial_\mu U=i\tau_a\partial_\mu\pi_a/F+\cdots$. Hence
$$
\mathcal L_{\pi}=\frac12\partial^\mu\boldsymbol\pi\cdot\partial_\mu\boldsymbol\pi+\cdots,
$$
so the three fields have canonical <kinetic terms>. There is no nonconstant invariant potential in the exact chiral theory, consistent with their masslessness.
For real diagonal $M$, the two terms in the mass trace are
$$
\operatorname{tr}(MV)=m_u\bar u_Lu_R+m_d\bar d_Ld_R,\qquad \operatorname{tr}(MV^\dagger)=m_u\bar u_Ru_L+m_d\bar d_Rd_L.
$$
Their sum therefore gives the usual <Dirac mass> term,
$$
\mathcal L_m=-m_u\bar uu-m_d\bar dd.
$$
These terms explicitly break <chiral symmetry>. In the low-energy <chiral perturbation theory> prescription, $V\mapsto-vU$ yields
$$
\mathcal L_{\pi,m}=v\operatorname{tr}[M(U+U^\dagger)].
$$
The <Pauli matrix> algebra implies $(\boldsymbol\pi\cdot\boldsymbol\tau)^2=|\boldsymbol\pi|^2I$. Thus
$$
U+U^\dagger=2I-\frac{|\boldsymbol\pi|^2}{F^2}I+O(\pi^4),
$$
and
$$
\mathcal L_{\pi,m}=2v(m_u+m_d)-\frac{v(m_u+m_d)}{F^2}|\boldsymbol\pi|^2+O(\pi^4).
$$
Comparing with the canonically normalized scalar mass term gives the <leading pion mass from a two-flavour condensate>:
$$
\boxed{m_{\pi_1}^2=m_{\pi_2}^2=m_{\pi_3}^2=\frac{2v(m_u+m_d)}{F^2}.}
$$
Positive quark masses therefore give positive pion squared masses. Even if $m_u\ne m_d$, this leading action depends only on their sum and gives equal masses for all three <pions>. Strong isospin-breaking effects beyond this order and electromagnetic effects can split them. In the condensate convention used here, $\langle\bar uu\rangle=\langle\bar dd\rangle=-2v$, which also explains the factor of two in the mass relation. The massive pions are <pseudo-Goldstone bosons>, becoming massless as both quark masses tend to zero.
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