Consecutive integers divisible by prime powers 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?

Hmm...
Well, by CRT $n \equiv 1 \pmod {p^2}$
$n \equiv 0 \pmod {q^3}$
$n \equiv-1 \pmod {r^4}$ will have unique solutions for any the primes $p, q, r$.
If I blanketly choose $r=2; q=3;p=5$ I will get.
$n\equiv 1 \pmod {25}$ and $n\equiv 0 \pmod 27$.  $n= 27j = 1 + 25k = 1-2k + 27k$.  If $k=14$ we have $j= 13$.  This gives us $n\equiv 351 \pmod {675}$.  
$n \equiv 351 \pmod{675}$ and $n\equiv -1 \pmod {16}$ gives us 
$351+ 675m\equiv  -1 + 3m \equiv -1 \pmod{16}$.
so $n = 351$ gives us $n-1 = 14*5^2; n = 13*3^2; n+1 =44*2^4$.
Is that the least possible?  
Well $3^4< 352 <5^4$ so $r=2,3$ if we are to find any lower.
And $7^3< 351 < 11^3$ so $q=3,5,7$ if we are to find any lower.
And $17^2 < 351 < 19^2$ so $p = 7,11,13,17$ if we are to find any lower.
If worse comes to worse we can solve them all.... Or we could write a program to check all numbers less than $351$.
I doubt there is any lower but I'm not betting anything valuable on it.
