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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

$aequiv 0pmodd^2$

$a+1equiv 0pmode^3$

$a+2equiv 0pmodf^4$

where $d,e,f$ are prime. This is equivalent to

$aequiv 0pmodd^2$

$aequiv -1pmode^3$

$aequiv -2pmodf^4$

Then after applying the Chinese Remainder Theorem,

$e^3f^4aequiv 1pmodd^2$

$d^2f^4aequiv 1pmode^3$

$d^2e^3aequiv 1pmodf^4$

I don't think this is getting me anywhere. Is this sort of general approach going to work or am I wasting my time?

Hint: It's a good idea to think about which primes these primes could be. For example, since you're looking for small numbers, you probably want to see what happens if you stipulate that the prime $p$ so $p^4$ divides your largest number is $2$ -- then try for $3$, $5$, etc. until you find that your result ends up being too big. I'll show you a worked example for the simpler problem of finding two consecutive numbers the first of which has a perfect square of a prime as a factor and the second of which has the perfect cube of a prime as a factor:

Let our integers be $a$ and $a+1$. We know that either $8|a+1$ (so $2$ is our prime) or $a+1geq 27$. Let's first try the case $8|a+1$. Then we want to find a number that the perfect square of a prime divides in the sequence

$$7, 15, 23, 31, 39, 47, 55, 63, 71...$$

and we find that the first one occurs at $63$, so ($63,64$) is a solution. Now let's try when $27|a+1$. We get that $a$ is in

$$26, 53, 80...$$

and then everything above $53$ is too big, since we've already found the example $(63,64)$. Now, if our prime is any bigger than $3$, we get that $a+1geq 125$, which is going to give us something much larger than $(63,64)$, and thus $(63,64)$ is our answer.

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.

$nequiv 1 pmod 25$ and $nequiv 0 pmod 27$. $n= 27j = 1 + 25k = 1-2k + 27k$. If $k=14$ we have $j= 13$. This gives us $nequiv 351 pmod 675$.

$n equiv 351 pmod675$ and $nequiv -1 pmod 16$ gives us

$351+ 675mequiv -1 + 3m equiv -1 pmod16$.

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.

I wrote a Python code to try to find these $3$ consecutive numbers, and I searched up to $100$ million or $10^8$, and couldn't find them. So if they exist, they're definitely larger than $10^8$.

If you want to continue the search, here is my code (it's for Python 3.7, to be exact, but it should work for Python 3.0+):

import math

lookup_range = 100000000

def isPrime(a):

 if(a <= 1):
 return False

 for i in range(2,a-1):
 if a % i == 0:
 return False
 return True

R = int(math.sqrt(lookup_range+1))+1

primelist = [a for a in range(R) if isPrime(a)]

for a in range(1,lookup_range+1):

 if(a % 1000000 == 0):
 print('Looked up to ' + str(a) + ', ' + str(round(100*a/lookup_range,2)) + '%')

 passedTestOne = False
 passedTestTwo = False
 passedTestThree = False

 p1 = 0
 p2 = 0
 p3 = 0

 for i in primelist:
 if(a % pow(i,2) == 0):
 passedTestOne = True
 p1 = i
 else:
 break

 for i in primelist:
 if((a+1) % pow(i,3) == 0):
 passedTestTwo = True
 p2 = i
 else:
 break

 for i in primelist:
 if((a+2) % pow(i,4) == 0):
 passedTestThree = True
 p3 = i
 else:
 break

 if(passedTestOne and passedTestTwo and passedTestThree):

 print(str(a) + ' = ' + str(p1) + '^2')
 print(str(a+1) + ' = ' + str(p2) + '^3')
 print(str(a+2) + ' = ' + str(p3) + '^4')
 break