The sum of the squares of two consecutive integers is 41. Find the integers

• Let x, (x+1) be the 2 integers
  
  x^2 + (x+1)^2 = 41
  x^2 + x^2 + 2x + 1 = 41
  2x^2 + 2x - 40 = 0
  x^2 + x - 20 = 0
  (x+5) (x-4) = 0
  x=-5 or x=4
  If x=-5, x+1= -4
  If x=4, x+1=5
  
  So the 2 integers are (4,5) or (-5,-4)
  
  
• 10 years agoHelpfull: Yes(3) No(0)
• x^2+(x+1)^2=41
  2x^2+2x-40=0
  x^2+x-20=0
  x=4 & x=-5
  thus 4,5 & -5,-4
• 10 years agoHelpfull: Yes(0) No(0)