Solving Project Euler: Problem 45 (opens in new tab)

I attempted the problem myself before reading your solution. My strategy is bit different: for every N>285, I check whether there exists a positive integer solution n to two equations T_N=P_n and T_N=H_n. Using basic algebra and quadratic formula, it boils down to checking whether the quantities 1+12(NN+N) and 1+4(NN+N) are perfect squares. If they are perfect squares, denote their squares by x and y. The next step is to check if 1+x is divisible by 6 and if 1+y is divisible by 4. They are easy using % operator.

    from math import sqrt

    def isPerfectSquare(n):
        sr = int(sqrt(n))
        return sr * sr == n

    for N in range(1, 10000000000):
        foo = 1 + 12*(N*N + N)
        bar = 1 + 4*(N*N + N)
        if isPerfectSquare(foo) and isPerfectSquare(bar):
            x, y = int(sqrt(foo)), int(sqrt(bar))
            if (1+x)%6 == 0 and (1+y)%4 == 0:
                print(f'Candidate found: N={N}, T_N={N*(N+1)/2}')

I just did a dumb search with lazy lists in Haskell, inspired by the famous problem about Hamming numbers. Runtime was 0.02 seconds. About 2 more seconds to fine the one after that (results: [1,40755,1533776805,57722156241751]).

    import           Data.Ord

    tri n = n*(n+1)`quot`2
    pent n = n*(3*n-1)`quot`2
    hex n = n*(2*n-1)

    fs :: (Integer -> Integer)->[Integer]
    fs f = map f [1..]

    cm aas@(a:as) bbs@(b:bs)
       = case compare a b of
           EQ -> a : cm as bs
           GT -> cm aas bs
           LT -> cm as bbs

    tt = (fs tri) `cm` (fs pent) `cm` (fs hex)

    main = print . take 3 $ tt

Added: the one after that is 2172315626468283465 which took about 6 minutes with the same Haskell program. I'm sure there are faster implementations and better algorithms. I didn't try to search any further.

The mask hides the whole sentence except the answer. Lol.