Question 398436
You want to find if there is some Pythagorean triple {x, x+2, x+4}, with {{{x <> 6}}}. Since x+4 is the longest side,


{{{x^2 + (x+2)^2 = (x+4)^2}}}


{{{x^2 + (x^2 + 4x + 4) = x^2 + 8x + 16}}} Moving everything to one side,


{{{x^2 - 4x - 12 = 0}}}


{{{(x - 6)(x + 2) = 0}}} --> x = 6 or x = -2. However, x cannot be negative, so the only triple is when x = 6, and they will never be Pythagorean triples again.