Question 28610
Let the 5 consecutive integers be represented as 

x, x+1, x+2, x+3, x+4

According to the problem, 

x^2 + (x+1)^2 + (x+2)^2 = (x+3)^2 + (x+4)^2 

expanding each term on both the right and left side of the equation
using the formula (a + b)^2 = a^2 + 2*a*b + b^2 

x^2 + x^2 + 2*x + 1 + x^2 + 4*x + 4 = x^2 + 6*x + 9 + x^2 + 8x + 16

grouping all the x^2 terms,  x terms and the constants on both the right and left side of the equation we get

3*x^2 + 6*x + 5 = 2x^2  + 14*x + 25

rewriting this equation as 
3x^2 - 2x^2 + 6*x - 14*x + 5-25 = 0

x^2 - 8*x -20 = 0 
we can use the quadratic formula 
x = (-b +/- sqrt(b^2 - 4*a*c)) /(2*a)
or factorise the equation  x^2 - 8*x -20 = 0 as (x-10)(x+2) = 0

x = 10 or x =-2

 The problems defines that consecutive positive integers

which means x = -2 is not a valid answer
Thus x =10
and the 5 consecutive positive integres are 10,11,12,13,14