When an AP makes a perfect square

Let the positive integers be

$\{ a \ , a+1 \ , a+2 \ , a+3 \ , a+4 \ , a+5 \}$

The sum of all these integers is

$S = \dfrac 62 \left( a + a+ 5 \right) = 3 \left( 2a + 5 \right)$

Let $S = n^2$ for some positive integer $n$ .

Thus we have

$n^2 = 3 \left( 2a + 5 \right) \implies \dfrac{n^2}{3} = 2a+5$

Since $a$ is an integer so $2a + 5$ is also an integer. This means that $n^2$ must be divisible by 3.

Hence, $n^2$ must be divisible by 9 (since we can't have a single factor of 3 in a perfect square). This implies that $n$ must be divisible by 3.

Let $n=3k$ for some positive integer $k < n$ .

Thus, we have

${(3k)}^2 = 3 (2a+5) \\ \implies 3k^2 = 2a + 5 \\ \implies a = \dfrac{3k^2 - 5}{2}$

Hence, for $a$ to be a positive integer $k$ should be an odd positive integer and $k \ge 3$ .

Thus, we derive that there can exist infinitely many perfect squares that satisfy the criterion which are of the form $9 {(2m+1)}^2$ where $m$ is any positive integer.

Let the first positive integer be $a$ , then the sum of six consecutive positive integers is $S = \dfrac 62 (a+a+5) = 6a + 15$ , which is divisible by 3. Then, we can assume:

$\begin{aligned} 6a+15 & = (3k)^2 & \small \color{#3D99F6} \text{where }k \in \mathbb N \\ 2a + 5 &= 3k^2 & \small \color{#3D99F6} \text{Since LHS is odd, } k \text{ is odd.} \\ 2a + 5 & = 3(2n-1)^2 & \small \color{#3D99F6} \text{Let } k=2n-1 \\ 2a + 5 & = 3(4n^2-4n+1) \\ & = 12n^2-12n+3 \\ \implies a & = 6n(n-1)-1 \end{aligned}$

We note that for $n \ge 2$ , $a = 11, 35, 71, ...$ infinitely many of them.