Question 245451
let n= the first integer n+1= the second integer and n+2= the  third integer
n^2+(n+1)^2=(n+2)^


work out the squares
(n+1)^2=n^2+2n+1
(n+2)^2=n^2+4n+4
n^2+ n^2+2n+1=n^2+4n+4
subtract n^2 from both sides
n^2+2n+1=+4n+4
subtract 4n from both sides

n^2-2n+1=4
subtract 4 from both sides
n^2-2n-3=0
n=3 and -1
try 3, 4 and 5
3^2+4^2=5^2
9+16=25
ok
try -1,0,1
(-1)^2+0^2=1^2
1=1
ok