Find the Sum of k when $k\in N$ and $\sqrt {k^2+48k} \in N$

[Albert1]

$k\in N$ , and $\sqrt {k^2+48k} $ $\in N$

find $\sum k$

 

[kaliprasad]

Spoiler

let $\sqrt{k^2+48k} = n$
so $k^2+48k=n^2$
or $(k+24)^2 -n^2= 576\cdots(1)$
or $(k+24+n)(k+24-n) = 576$
further from (1) both (k+24) and n have to be even or odd so (k+24+n) and (k+ 24-n) both are even and k+24+n > 24
so we get
$(k+24+n, k+24-n) = (288,2)$ giving $k = 121$
or $(144,4)$ giving $k= 50$
or $(72,8)$ giving $k=16$
or$(48,12)$ giving $k=6$
or $(36,16)$ giving $k =2$
or $(32,18)$ giving $k=1$
so sum of $k = 1 + 2 + 6+16 +50+121= 196$

edit: I had missed a solution
I missed (96,6) giving k= 27 giving sum of k = 223.

 

[Albert1]

Spoiler

let $\sqrt{k^2+48k} = n$
so $k^2+48k=n^2$
or $(k+24)^2 -n^2= 576\cdots(1)$
or $(k+24+n)(k+24-n) = 576$
further from (1) both (k+24) and n have to be even or odd so (k+24+n) and (k+ 24-n) both are even and k+24+n > 24
so we get
$(k+24+n, k+24-n) = (288,2)$ giving $k = 121$
or $(144,4)$ giving $k= 50$
or $(72,8)$ giving $k=16$
or$(48,12)$ giving $k=6$
or $(36,16)$ giving $k =2$
or $(32,18)$ giving $k=1$
so sum of $k = 1 + 2 + 6+16 +50+121= 196$

thanks for participation , but your answer is not correct,there is one answer missing

 

[kaliprasad]

thanks for participation , but your answer is not correct,there is one answer missing

Spoiler

Yes I missed (96,6) giving k= 27 giving sum of k = 221.
Note I shall edit the post also

 

[Albert1]

Spoiler

Yes I missed (96,6) giving k= 27 giving sum of k = 221.
Note I shall edit the post also

Spoiler

sum of k=223

 

[kaliprasad]

Spoiler

sum of k=223

OOPS one more mistake