Existence of positive integer k that are both squares Is there a positive integer k such that $4k+1$ and $9k+1$ are both squares?

The answer is no for k>0.
Let $b>a$, $a^2=4k+1$ and $b^2=9k+1$. Note that, 
$$
5k=b^2-a^2=(b+a)(b-a)=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)\cdot \left(\sqrt{9k+1}-\sqrt{4k+1}   \right)
$$
Case 1:set $k$ a prime number. Then


*

*$k=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$ and  $5=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ or 

*$k=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ and $5=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$. 
In both cases, 
$
k+5=2\sqrt{9k+1}.
$
And this quadratic equation has no integer solution to $ k $.
Case 2:set $k=\alpha\cdot\beta\cdot p$ whit $\beta\geq \alpha$ and $p>1$ natural numbers. By a procedure completely analogous to the previous case we obtain,


*

*$\beta\cdot p =\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$ and  $5\alpha=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ or 

*$\beta\cdot p=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ and $5\alpha=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$. 
In both cases, $p\beta +5\alpha=2\sqrt{9\alpha\beta p+1}.$ Also in this case you can also check that there is no real solution to $ p $ for all $\alpha,\beta\in\mathbb{N}^*$.
