Question 534044
Let x be the first integer
Then x+1 be the second
And x+2 be the third


{{{(x+1)(x+2)=3(x+1)+3}}}
{{{x^2+2x+x+2=3x+3+3}}}
{{{x^2+3x+2=3x+6}}}
{{{x^2+3x-3x=6-2}}}
{{{x^2=4}}}
{{{sqrt(x^2)=sqrt(4)}}}
x=+-2 ---Sorry can't enter the +_ Sign


So the first integer is either 2 or -2


The second integer
=+-2+1
=3/-1


The third integer
=+-3+1
=4/0


Filtering out:
3(4)=3(3)+3
12=9+3
12=12
TRUE


3(-1)=3(0)+3
-3=0+3
-3=3
FALSE


So the three integers are 3, 4 and 5