Prime Power Triples

May 22, 2020

Today’s exercise is Project Euler Problem 87:

The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:

28 = 22 + 23 + 24
33 = 32 + 23 + 24
49 = 52 + 23 + 24
47 = 22 + 33 + 24

How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?

Your task is to solve Project Euler 87; in the spirit of Project Euler, show only your code but not the solution. When you are finished, you are welcome to read
or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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5 Responses to “Prime Power Triples”

  1. Zack said

    Nifty little drill. Here is my take on it using Julia 1.4.1:

    The original approach I took was way too sluggish, so I ended up using a slicker one. There is also a verbose option for viewing the specific combinations: main(n, true), where n is the maximum integer to explore (in this case 50 000 000). The default for this parameter is false.

    Glad the blog is back! Cheers

  2. Jan Van lent said
    from sympy import primerange
    def count(N):
        return len(set(i**2 + j**3 + k**4
                   for i in primerange(1, int(N**(1/2))+1)
                   for j in primerange(1, int((N-i**2)**(1/3))+1)
                   for k in primerange(1, int((N-i**2-j**3)**(1/4))+1)))
  3. Richard A. O'Keefe said

    Easy to solve by brute force, once you realise that there are not that many prime powers to look at.

  4. matthew said

    Welcome back! Hope all is well with everybody.

    Not sure I like those multiple calls to primerange etc. Here’s another Python solution that generates a single list of prime numbers. Takes a couple of seconds for the Euler problem (here we just count up to a million):

    # Recursive sieve of Eratosthenes
    # Squares the size of the max prime each time
    def rsieve(primes):
        N = primes[-1]**2+1
        sieve = [0]*N
        for p in primes:
            for q in range(p*p,N,p):
                sieve[q] = 1
        return [i for i in range(2,N) if sieve[i] == 0]
    N = 1000000
    primes = [2]
    while primes[-1]**2 < N:
        primes = rsieve(primes)
    s = set()
    for p in primes:
        ptotal = p**2
        if ptotal > N: break
        for q in primes:
            qtotal = ptotal+q**3
            if qtotal > N: break
            for r in primes:
                rtotal = qtotal+r**4
                if rtotal > N: break
  5. Alex B said

    Very glad to hear you are well and that things are settling down somewhat.

    Here’s my solution in Python, using a standard Sieve of Eratosthenes and precalculating the required prime powers. Runs in about half a second on my desktop.

    #! /usr/bin/env python3
    # Euler 87
    # Also Programming Praxis
    # Find count of numbers n < 50e6 where n = p^2 + q^3 + r^4; p, q, r all prime
    # Limits:
    #   ceil (N ** 1/2) = 7072
    #   ceil (N ** 1/3) = 369
    #   ceil (N ** 1/4) = 85
    import time
    import primes
    start = time.time()
    # Eratosthenes sieve primes up to largest limit
    ps = primes.sieve(7072)
    # Precalculate prime powers possible
    p2s = [p*p for p in ps]
    p3s = [p*p*p for p in ps if p <= 369]
    p4s = [p*p*p*p for p in ps if p <= 85]
    # add and collect unique answers
    sums = set()
    lim = 50_000_000
    for p2 in p2s:
        for p3 in p3s:
            for p4 in p4s:
                tot = p2 + p3 + p4
                if tot < lim:
    end = time.time()
    taken = end - start
    print(f'Calculated in {taken:.3} seconds')

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