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

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SUMMARY

The discussion focuses on finding the sum of natural numbers \( k \) such that \( \sqrt{k^2 + 48k} \) is also a natural number. The correct approach involves recognizing that \( k^2 + 48k \) must be a perfect square, leading to the equation \( n^2 - k^2 = 48k \) for some integer \( n \). The solutions yield specific values of \( k \), which when summed provide the final answer.

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

find $\sum k$
 
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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.
 
Last edited:
kaliprasad said:
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
 
Albert said:
thanks for participation , but your answer is not correct,there is one answer missing

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

sum of k=223
 
Last edited by a moderator:
Albert said:
sum of k=223

OOPS one more mistake
 

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