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?
 A: Let's find the $k$ for which $4k+1 = a^2$ is a square. $4k+1$ is odd, so $a$ is odd. Then we have
$$k = \frac{a^2-1}{4} = \frac{a-1}{2}\cdot\frac{a+1}{2} = \frac{a-1}{2}\left(\frac{a-1}{2}+1\right)$$
the product of two consecutive numbers. For convenience, let's write $k =m\cdot (m+1)$. So can $9k+1$ be a square then?
$$9k+1 = 9m(m+1)+1 = 9m^2 + 9m + 1 = (3m+1)^2 + 3m = (3m+2)^2 - 3(m+1)$$
lies strictly between two consecutive squares then, unless $m = 0$ or $m+1 = 0$, which means $k = 0$, but that is excluded since $k$ was supposed to be positive.
A: 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.
A: 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 $\alpha,\beta\in\mathbb{N}^*$.
