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?

Suppose that $4k+1 = a^2$ and $9k+1 = b^2$, where $a, b$ are positive integers.
Then, we get that $ 9a^2 - 4b^2 = 5 $.
Since we have $(3a-2b)(3a+2b) = 5$, and both terms are integers,$3a+2b > 0$, $3a+2b>3a-2b$, so we must have $3a+2b = 5, 3a-2b = 1$. This gives us $a=1, b= 1$ and hence $k=0$ is the only solution.
