<PRE><B>Question 57584</B>
how do i find three consecutive intergers such that the sum of the squares is 77.
Let the consecutive integers be x, x+1, x+2.
EQUATION:
x^2+(x^1)^2+(x+2)^2=77
x^2+x^2+2x+1 + x^2+4x+2=77
3x^2+6x+3=77
3x^2+6x-74=0
x=[-6+-sqrt(6^2-4*3*-74)]/6
x=[-6+-sqrt(924)]/6
Comment: This answer is not an integer.
There are no consecutive integers that meet
the requirement of the equation.
Cheers,
Stan H.</PRE>