Question 288406
Recall that consecutive integers follow the pattern: x, x+1, x+2, ...


So the statement "3 consecutive integers such that the sum of the squares of the smaller two is equal to the square of the largest" means that {{{x^2+(x+1)^2=(x+2)^2}}}



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



{{{x^2+x^2+2x+1=x^2+4x+4}}} <a href="http://www.algebra.com/algebra/homework/Distributive-associative-commutative-properties/foil.solver">FOIL</a>



{{{2x^2+2x+1=x^2+4x+4}}} Combine like terms.



{{{2x^2+2x+1-x^2-4x-4=0}}} Get every term to the left side.



{{{x^2-2x-3=0}}} Combine like terms.



Notice that the quadratic {{{x^2-2x-3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-2}}}, and {{{C=-3}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-2) +- sqrt( (-2)^2-4(1)(-3) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-2}}}, and {{{C=-3}}}



{{{x = (2 +- sqrt( (-2)^2-4(1)(-3) ))/(2(1))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{x = (2 +- sqrt( 4-4(1)(-3) ))/(2(1))}}} Square {{{-2}}} to get {{{4}}}. 



{{{x = (2 +- sqrt( 4--12 ))/(2(1))}}} Multiply {{{4(1)(-3)}}} to get {{{-12}}}



{{{x = (2 +- sqrt( 4+12 ))/(2(1))}}} Rewrite {{{sqrt(4--12)}}} as {{{sqrt(4+12)}}}



{{{x = (2 +- sqrt( 16 ))/(2(1))}}} Add {{{4}}} to {{{12}}} to get {{{16}}}



{{{x = (2 +- sqrt( 16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (2 +- 4)/(2)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{x = (2 + 4)/(2)}}} or {{{x = (2 - 4)/(2)}}} Break up the expression. 



{{{x = (6)/(2)}}} or {{{x =  (-2)/(2)}}} Combine like terms. 



{{{x = 3}}} or {{{x = -1}}} Simplify. 



So the solutions are {{{x = 3}}} or {{{x = -1}}} 

  

This means that the numbers are either: 3, 4, 5


OR


the numbers are: -1, 0, 1