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
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