We have:```(1) e1 * d = 1 (mod phi)(2) e2 * (d + 2) = 1 (mod phi)```Multiply (2) by e1 and (1) by e2 and subtract them:```2 * e1 * e2 = e1 - e2 (mod phi)```That means `2 * e1 * e2 - (e1 - e2)` is a multiple of `phi`, and can be used to calculate `d`, like what we normally do:```d = inverse(e1, 2 * e1 * e2 - (e1 - e2))```Now you know the rest:```pt = pow(c, d, n)print(long_to_bytes(pt))```Flag: `TWCTF{even_if_it_is_f4+e}`