Note the following identity, \begin{align*} (a^2 + 6b^2)(c^2 + 6d^2) = (ac+6bd)^2+6(da-bc)^2 \end{align*}So, this means that multiplying two expressions of the type $u^2+6v^2$ gives another expression of this type. Notice that $1^2+6\cdot 1^2 = 7$. Because of these two, inductively, we know that there exist some integers $x$ and $y$ such that $x^2+6y^2=7^k$. Remark: The identity above is true for any $Q_k(x,y)=x^2 + ky^2$, not only for $k = 6$. Tags: Bivariate quadratic forms, Induction