Question 81771
Let x=1st #, y=2nd #, z=3rd #

Since you have consecutive integers, this means the integers are right next to each other. For instance, 1,2, and 3 are all consecutive integers. If x=1, then y=2, and z=3. So if you wanted them all in terms of x, you would say:

x=1, y=x+1, z=x+2


Since if you plug in x=1 into x+2, you get z=3. So the same applies to this problem. 


So now we're adding the squares of these numbers to get 77 like this:


{{{x^2+y^2+z^2=77}}}

But we can write y as {{{x+1}}} and z as {{{x+2}}} so we can replace y and z:


{{{x^2+(x+1)^2+(x+2)^2=77}}}



{{{x^2+x^2+2x+1+x^2+4x+4=77}}} Foil the two parenthesis separately


{{{3x^2+6x+5=77}}} Combine like terms


{{{3x^2+6x+5-77=cross(77-77)}}} Subtract 77 from both sides


{{{3x^2+6x-72=0}}}


Now that we have a quadratic, lets use the quadratic formula to solve for x:

*[invoke quadratic "x", 3, 6, -72 ]


So we have 2 possible numbers for our first number. 
Now lets check our first possible answer x=4:

Lets find our 2nd number using x=4:

{{{y=x+1}}}

{{{y=4+1}}} Plug in x=4

{{{y=5}}}

So our 2nd number is 5

Lets find our 3rd number using x=4

{{{z=x+2}}}

{{{z=4+2}}} Plug in x=4

{{{z=6}}}

So our 3rd number is 6

So our 3 numbers are 4,5,6


Check:

{{{4^2+5^2+6^2=77}}} Plug in x=4,y=5,z=6

{{{16+25+36=77}}}

{{{77=77}}} works




Now lets check our other possible answer x=-6:

Lets find the 2nd number using x=-6

{{{y=x+1}}}

{{{y=-6+1}}} Plug in x=-6

{{{y=-5}}}

So our 2nd number is -5

Lets find our 3rd number using x=-6

{{{z=x+2}}}

{{{z=-6+2}}} Plug in x=-6

{{{z=-4}}}

So our 3rd number is -4

So our 3 numbers are -6,-5,-4


Check:

{{{(-6)^2+(-5)^2+(-4)^2=77}}} Plug in x=-6,y=-5,z=-4

{{{36+25+16=77}}}

{{{77=77}}} works


So we have 2 sets of numbers that work. We have one set


4,5,6  


and we also have


-6,-5,-4