The K One

We note that $k(k+15)$ is always even. Therefore, we can write $k(k+15) = (2n)^2$ , where $n$ is a positive integer. Then, we have:

$\begin{aligned} k(k+15) & = 4n^2 \\ k^2 + 15k - 4n^2 & = 0 & \small \color{#3D99F6} \text{Solving the quadratic for }k \\ \implies k & = \dfrac {\sqrt{15^2+(4n)^2}-15}2 \end{aligned}$

For $k$ to be an integer, $15^2+(4n)^2$ must be an odd perfect square. This means that $15$ and $4n$ are the smaller of a Pythagorean triple . The relevant Pythagorean triples are:

$\begin{array} {cll} {\color{#3D99F6}8}, \underline{15}, \color{#D61F06}17 & \implies \color{#3D99F6} n=\dfrac 84 = 2 & \implies k = \dfrac {{\color{#D61F06}17}-15}2 = 1 \\ \underline{15}, {\color{#3D99F6}20}, \color{#D61F06} 25 & \implies \color{#3D99F6} n= \dfrac {20}4 = 5 & \implies k = \dfrac {{\color{#D61F06}25}-15}2 = 5 \\ \underline{15}, {\color{#3D99F6}36}, \color{#D61F06} 39 & \implies \color{#3D99F6} n= \dfrac {36}4 = 9 & \implies k = \dfrac {{\color{#D61F06}39}-15}2 = 12 \\ \underline{15}, {\color{#3D99F6}112}, \color{#D61F06} 113 & \implies \color{#3D99F6} n= \dfrac {112}4 = 28 & \implies k = \dfrac {{\color{#D61F06}113}-15}2 = 49 \end{array}$

Therefore, the sum of all possible $k$ s is $1+5+12+49=\boxed{67}$ .

We require that $k(k + 15) = n^{2}$ for some positive integer $n$ . This equation can be written as

$k^{2} + 15k = n^{2} \Longrightarrow \left(k + \dfrac{15}{2} \right)^{2} - \dfrac{225}{4} = n^{2} \Longrightarrow (2k + 15)^{2} - 225 = 4n^{2}$

$\Longrightarrow (2k + 15)^{2} - (2n)^{2} = 225 \Longrightarrow ((2k + 15) + 2n)((2k + 15) - 2n) = 225$ .

Now consider the factorization pairs of $225 = 3^{2}5^{2}$ , ascribing the largest factor to $(2k + 15) + 2n$ and the smallest to $(2k + 15) - 2n$ , and then solving for $k$ . There are 5 such factorizations:

• $(2k + 15) + 2n = 225, (2k + 15 - 2n = 1 \Longrightarrow 4k + 30 = 226 \Longrightarrow k = 49$ ,

• $(2k + 15) + 2n = 75, (2k + 15) - 2n = 3 \Longrightarrow 4k + 30 = 78 \Longrightarrow k = 12$ ,

• $(2k + 15) + 2n = 45, (2k + 15) - 2n = 5 \Longrightarrow 4k + 30 = 50 \Longrightarrow k = 5$ ,

• $(2k + 15) + 2n = 25, (2k + 15) - 2n = 9 \Longrightarrow 4k + 30 = 34 \Longrightarrow k = 1$ ,

• $(2k + 15) + 2n = 15, (2k + 15) - 2n = 15 \Longrightarrow 4k + 30 = 30 \Longrightarrow k = 0$ .

The desired sum of positive integers $k$ is then $49 + 12 + 5 + 1 + 0 = \boxed{67}$ .

The key fact we use here repeatedly is that if $u$ and $v$ are coprime integers and $uv$ is a square number, then each of $u$ and $v$ is also a square number (this is easy to prove - just consider occurrences of prime factors).

The two factors $k$ and $k+15$ may or may not be coprime. Indeed, $\gcd{(k,k+15)}=\gcd{(k,15)} \in \{1,3,5,15\}$ . We need to look at each of these cases separately.

Case $\gcd{(k,k+15)}=1$ :

$k=a^2$ , $k+15=b^2$

$b^2-a^2=(b-a)(b+a)=15$

There are two sub-cases: $b-a=1$ , $b+a=15$ : $a=7,b=8$ ; $\boxed{k=49}$

$b-a=3$ , $b+a=5$ : $a=1,b=4$ ; $\boxed{k=1}$

This is the only case where the two factors are automatically coprime. For the other cases, we need to divide through by the common factor first.

Case $\gcd{(k,k+15)}=3$ :

$k=3k'$

$k'=a^2$ , $k'+5=b^2$

$b^2-a^2=(b-a)(b+a)=5$

$b-a=1$ , $b+a=5$ : $a=2,b=3$ ; $\boxed{k=12}$ .

Case $\gcd{(k,k+15)}=5$ :

$k=5k'$

$k'=a^2$ , $k'+3=b^2$

$b^2-a^2=(b-a)(b+a)=3$

$b-a=1$ , $b+a=3$ : $a=1,b=2$ ; $\boxed{k=5}$ .

Case $\gcd{(k,k+15)}=15$ :

$k=15k'$

$k'=a^2$ , $k'+1=b^2$

But there are no consecutive positive square numbers; so no solutions in positive integers for $k$ in this case.

Hence the full solution set for $k$ is $\{1,5,12,49\}$ , with sum $\boxed{67}$ .

Let a be a positive integer such that k^2+15k=a^2. Then 4a^2+225 must be a perfect square, say it is equal to b^2 where b is another positive integer. Then (b+2a)(b-2a)=225. Writing 225 as the product of 1 and 225, 3 and 75, 5 and 45, and 9 and 25 (the option 225=15.15 gives b=0 which makes k negative). We get the values of k as 49,12,5,1 whose sum is 67