Question 81771
The first thing to do is to let x be the first integer. Then the next consecutive integer
is 1 number higher or x + 1. And the next consecutive integer after that is (x+1) + 1 or
x + 2.
.
The squares of these numbers are:
.
{{{(x^2)}}}
.
{{{(x + 1)^2 = x^2 + 2x + 1}}}
.
{{{(x + 2)^2 = x^2 + 4x + 4}}}
.
If you add these three squares together you get:
.
{{{x^2 + x^2 + 2x + 1 + x^2 + 4x + 4}}}
.
Combine the like terms and you get:
.
{{{3x^2 +(2 + 4)x + (1 + 4)}}}
.
This reduces to:
.
{{{3x^2 + 6x + 5}}}
.
This is the sum of the squares and the problem tells you that this sum equals 77. So
the equation becomes:
.
{{{3x^2 + 6x + 5 = 77}}}
.
Put this into standard quadratic form by subtracting 77 from both sides to get:
.
{{{3x^2 + 6x - 72 = 0}}}
.
Since this is in standard quadratic form, you can solve it by using the quadratic
formula. Although it may be difficult to recognize, the left side of this equation can
be factored:
.
{{{3(x - 4)(x + 6) = 0}}}
.
Divide both sides by 3 to reduce the equation to:
.
{{{(x - 4)(x + 6) = 0}}}
.
This equation will be true if one of the factors is zero because a multiplier of zero on the
left side will make the left side equal to zero, just as the right side is.
.
So solve this equation by setting each factor equal to zero.
.
{{{x - 4 = 0}}}
.
Solve by adding 4 to both sides to get {{{x = 4}}}. This would mean that the next two 
consecutive integers are 5 and 6.
.
Now set the second factor equal to zero:
.
{{{(x + 6) = 0}}}
.
Solve by subtracting 6 from both sides to get {{{x = - 6}}}.  This makes the next two
consecutive integers (x+1 and x + 2) equal to -5 and -4.
.
So we have two possibilities. One solution is 4, 5, and 6. The other solution is -6, -5,
and -4.
.
Let's check them out. 
.
{{{4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77}}}
.
That works. Now:
.
{{{(-6)^2 + (-5)^2 + (-4)^2 = 36 + 25 + 16 = 77}}}
.
Although it's a little more difficult to think about, -6, -5, and -4 are three consecutive
integers that also work.
.
Hope this helps you to understand the problem.