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

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The discussion revolves around finding the sum of natural numbers \( k \) such that \( \sqrt{k^2 + 48k} \) is also a natural number. Participants are attempting to solve the equation but indicate that some answers provided are incorrect and that one solution is missing. There are multiple corrections and clarifications made throughout the conversation. The focus remains on ensuring all valid solutions for \( k \) are identified and summed correctly. The thread emphasizes the importance of accuracy in solving the mathematical problem presented.
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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
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
View full post »

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