Question 195949
Consecutive integers follow the form: x, x+1, x+2, x+3, etc...



So...


"The sum of the squares of three consecutive, positive integers is equal to the sum of the squares of the next two integers." translates to {{{x^2+(x+1)^2+(x+2)^2=(x+3)^2+(x+4)^2}}}



{{{x^2+(x+1)^2+(x+2)^2=(x+3)^2+(x+4)^2}}} Start with the given equation.



{{{x^2+x^2+2x+1+x^2+4x+4=x^2+6x+9+x^2+8x+16}}} FOIL



{{{3x^2+6x+5=2x^2+14x+25}}} Combine like terms.



{{{3x^2+6x+5-2x^2-14x-25=0}}} Get all terms to the left side.



{{{x^2-8x-20=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-8}}}, and {{{c=-20}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-8) +- sqrt( (-8)^2-4(1)(-20) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-8}}}, and {{{c=-20}}}



{{{x = (8 +- sqrt( (-8)^2-4(1)(-20) ))/(2(1))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{x = (8 +- sqrt( 64-4(1)(-20) ))/(2(1))}}} Square {{{-8}}} to get {{{64}}}. 



{{{x = (8 +- sqrt( 64--80 ))/(2(1))}}} Multiply {{{4(1)(-20)}}} to get {{{-80}}}



{{{x = (8 +- sqrt( 64+80 ))/(2(1))}}} Rewrite {{{sqrt(64--80)}}} as {{{sqrt(64+80)}}}



{{{x = (8 +- sqrt( 144 ))/(2(1))}}} Add {{{64}}} to {{{80}}} to get {{{144}}}



{{{x = (8 +- sqrt( 144 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (8 +- 12)/(2)}}} Take the square root of {{{144}}} to get {{{12}}}. 



{{{x = (8 + 12)/(2)}}} or {{{x = (8 - 12)/(2)}}} Break up the expression. 



{{{x = (20)/(2)}}} or {{{x =  (-4)/(2)}}} Combine like terms. 



{{{x = 10}}} or {{{x = -2}}} Simplify. 



So the answers are {{{x = 10}}} or {{{x = -2}}} 

  

Once again, the problem mentions that the numbers are positive. So this means that the only solution is {{{x = 10}}}



This means that the numbers are: 10, 11, 12, 13, and 14