Question 582215
I know that the most classic Pythagorean triple is (3,4,5), but I will pretend that I do not know.
Let the three consecutive integers be x-1, x, and x+1.
If they are lengths of the three sides of a right triangle, the largest, x+1, is the length of the hypotenuse.
Pythagoras theorem says that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In this case,
{{{(x-1)^2+x^2=(x+1)^2}}} --> {{{x^2-2x+1+x^2=x^2+2x+1}}} --> {{{2x^2-2x+1=x^2+2x+1}}} --> {{{2x^2-2x+1-x^2-2x-1=x^2+2x+1-x^2-2x-1}}} --> {{{x^2-4x=0}}} --> {{{x(x-4)=0}}}
The solutions to that equation are {{{x=0}}} and {{{x=4}}}.
{{{x=0}}} does not make sense for the middle integer, because the length of the shortest side would be -1, and we expect those lengths to be positive.
So the middle integer is {{{highlight(x=4)}}} and the other two numbers are
{{{highlight(x-1=3)}}} and {{{highlight(x+1=5)}}}.
The lengths of the three sides of the right triangle are the consecutive integers 3, 4, and 5.