= Solution
Let $\bar X=n^{-1}\sum_iX_i$ and $S=\sum_i(X_i-\bar X)^2$. The <Gaussian distribution> <likelihood> is proportional to $\tau^{n/2}\exp[-\tau\sum_i(X_i-\mu)^2/2]$. Multiplying by the <normal-gamma distribution> prior gives the quadratic exponent $\sum_i(X_i-\mu)^2+K_0(\mu-\mu_0)^2+2\beta_0$. Completing the square,
$$
\sum_i(X_i-\mu)^2+K_0(\mu-\mu_0)^2=(K_0+n)(\mu-\mu_n)^2+S+\frac{K_0n}{K_0+n}(\bar X-\mu_0)^2.
$$
The power of $\tau$ is $\tau^{\alpha_0+n/2-1}\sqrt\tau$, so the <posterior distribution> has exactly the required family with
$$
\boxed{K_n=K_0+n,\qquad\mu_n=\frac{K_0\mu_0+n\bar X}{K_0+n},\qquad\alpha_n=\alpha_0+\frac n2,}
$$
and
$$
\boxed{\beta_n=\beta_0+\frac12\sum_i(X_i-\bar X)^2+\frac{K_0n}{2(K_0+n)}(\bar X-\mu_0)^2.}
$$
For a proper prior, $K_0,\alpha_0,\beta_0>0$ and $\mu_0\in\mathbb R$. The separate $\sqrt\tau$ factor is retained in both densities; it does not add $1/2$ to the update of $\alpha$. This is a <conjugate prior> for the Gaussian mean and <precision parameter>.
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