Question 869821: What are 5 consecutive integers such that the sum of squares of the greater two is equal to the sum of squares of the other three?
Found 2 solutions by Fombitz, htmentor:
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! Let the integers be N-2,N-1,N,N+1,N+2
So,
Two solutions:
Then the integers are {-2,-1,0,1,2}
and
Then the integers are {10,11,12,13,14}
Answer by htmentor(1343)  (Show Source):
You can put this solution on YOUR website! If we define our integers like this n-2,n-1,n,n+1,n+2, then most terms will cancel on either side of the equation when we square them
(n+1)^2 + (n+2)^2 = n^2 + (n-1)^2 + (n-2)^2
If you perform the multiplication, and collect terms you will be left with:
n^2 - 12n = 0
n(n-12) = 0
This gives two solutions, n=0 and n=12
So the integers are -2,-1,0,1,2 and 10,11,12,13,14
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