Question 430309
If we let the integers be x, x+1, and x+2, then


{{{x^2 + (x+1)^2 + (x+2)^2 = 3x^2 + 6x + 5}}}. This number is congruent to 2 modulo 3, since {{{3x^2 + 6x}}} is divisible by 3 and 5 is 2 modulo 3. Note that the perfect squares are either 0 or 1 modulo 3. This means that {{{3x^2 + 6x + 5}}} can never be equal to a perfect square.