Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals to the total of the three original integers. Which of the following best describes the minimum, say m of these three numbers?

Solution follows here:

Solution:

Let the numbers be n-1,n,n+1

(n-1)+n

^{2}+(n+1)^{3}= (n-1+n+n+1)^{2}=> n-1+n^{2}+n^{3}+1+3n+3n^{2}= (3n)^{2}
n

^{3}+4n^{2}+4n = 9n^{2}=> n^{3}-5n^{2}+4n = 0 => n(n^{2}-5n+4) = 0
n(n-1)(n-4) = 0 => n= 0 or 1 or 4

for n=0, the numbers are: -1,0,1
this is out as all the numbers should be positive

for n=1, the numbers are: 0,1,2
this is out as all the numbers should be positive (‘0’ can’t be taken as
positive)

for n=4, the numbers are: 3,4,5
this is correct and the minimum number out of the three is ‘3’

For ‘3’, the best option to select is (1)

**Answer (1)**

Awesome man! Thanks a tonne!

ReplyDeletecan you explain this part "(n-1)+n2+(n+1)3 = (n-1+n+n+1)2 => n-1+n2+n3+1+3n+3n2 = (3n)2"

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