Question 135065
You have more than a clue. Don't fret, You are part way there already.

Your solution uses the number n. But there is nothing that ensures that 'n' is even. So let's use a different set of numbers.
Let's assume n is even, if so, then n/2 is still an integer.
Our three numbers will be n/2, (n+4)/2 and (n+8)/2

Add in one more given " four times the middle integer is equal to two less than the sum of the first and third integers."

so {{{4*((n+4)/2) = ((n/2 + (n+8)/2)) - 2 )}}}
Left left side is four times the middle number. The right side is the (sum of the first and third) minus 2

Now solve

{{{4*((n+4)/2) = ((n/2 + (n+8)/2)) - 2 )}}}
{{{4*(n+4) = ((n + (n+8)) - 4 )}}}
{{{4n+16) = (2n + 8 - 4 )}}}
{{{4n+16) = 2n + 4 }}}
{{{2n = -12 }}}
n = -6

Check : is 4(-4) = (-6 + -2) - 2?
Nope, so I think there is not a solution to this one