SOLUTION: find three consecutive integers such that the sum of the squares of the smaller two is equal to the square of the largest.

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Question 18612: find three consecutive integers such that the sum of the squares of the smaller two is equal to the square of the largest.
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
find three consecutive integers such that the sum of the squares of the smaller two is equal to the square of the largest.


Let the integers be x,(x+1),(x+2)
Then squaring the smaller numbers we get
(x)^2+(x+1)^2=(x+2)^2
x^2+x^2+1+2x=x^2+4+4x
Simplyifying,
2x^2+2x+1=x^2+4x+4
Taking everything to one side,
2x^2-x^2+2x-4x+1-4=0
Which leaves us with,
x^2-2x-3=0
Factoring this quadratic we get,
(x-3)(x+1)=0
Therefore we get,
x=3,-1
Now taking values of x=3,-1 we get,
*** x=3,x+1=4,x+2=5
[3^2+4^2=9+16=25=5^2]
*** x=-1,x+1=0,x+2=1
[(-1)^2+(0)^2=1+0=1=(1)^2]


So our solution sets are:
(3,4,5) and (-1,0,1)


Hope this helps,
Prabhat