There's a similar question on here already but I don't think the answers are applicable in this case.
Question: Find the smallest three consecutive integers for which the first integer is divisible by the square of a prime; the second integer by the cube of a prime; and the third integer by the fourth power of a prime.
My attempt: Let the three consecutive integers be a, b and c. Then
$a\equiv 0\pmod{d^2}$
$a+1\equiv 0\pmod{e^3}$
$a+2\equiv 0\pmod{f^4}$
where $d,e,f$ are prime. This is equivalent to
$a\equiv 0\pmod{d^2}$
$a\equiv -1\pmod{e^3}$
$a\equiv -2\pmod{f^4}$
Then after applying the Chinese Remainder Theorem,
$e^3f^4a\equiv 1\pmod{d^2}$
$d^2f^4a\equiv 1\pmod{e^3}$
$d^2e^3a\equiv 1\pmod{f^4}$
I don't think this is getting me anywhere. Is this sort of general approach going to work or am I wasting my time?