$\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational Let $a,b \in \mathbb{N}^{*}$. Prove that $\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational.
If $(a,b)=(k,k\cdot6)$, then $\sqrt{13a^2+b^2}$ is rational, but I do not know if those are the only solutions. 
 A: If both are rational then $14(a^2+b^2)$ is the sum of two squares.  But it is impossible for (non-zero) $n$ and $14n$ to both be the sum of two squares as the well-known characterization requires the power of $7$ dividing each to be even.
A: $\sqrt{x}$ (where $x\in\mathbb Z^+$) is rational if and only if $\sqrt{x}$ is a positive integer (see this question).
Therefore, your problem is equivalent to proving the system $13a^2+b^2=k^2$ with $a^2+13b^2=m^2$ has no positive integer solutions.
$$a^2+13\left(k^2-13a^2\right)=m^2\iff m^2+168a^2=13k^2$$
I'll prove the last equation has no positive integer solutions.
$m^2\equiv 5k^2\pmod{8}$. But we have $y^2\equiv \{0,1,4\}\pmod{8}$ for all $y\in\mathbb Z$.
Therefore $m,k$ are both even. Let $m=2m_1$, $k=2k_1$. Then $m_1^2+42a^2=13k_1^2$.
$m_1,k_1$ have the same parity. If $m_1,k_1$ are both odd, then $m_1^2\equiv k_1^2\equiv 1\pmod{8}$, so $42a^2\equiv 4\pmod{8}$, i.e. $a^2\equiv 2\pmod{4}$, contradiction ($2$ is not a quadratic residue mod $4$). Therefore $m_1,k_1$ are both even, so $a$ is also even.
Therefore $(m,a,k)$ is a positive integer solution if and only if $(m/2,a/2,k/2)$ is also a positive integer solution. By Infinite Descent we get $m=a=k=0$.
