Square numbers as products of triangular numbers

James Tanton asks: “An old question: Triang nmbrs:1,3,6,10,15,.. For every triangular nmbr is there a second larger triang nmbr so that their product is square?”

Answer: yes! I thought “hey, this is a question I can answer in my head!” but didn’t get further than noticing that 1 times 36 is 36. Then I got home and wrote some Python. (You’ll probably see this blog’s language of choice shifting from R to Python, mostly because I’m using Python at work and this gives me a chance to practice.)

import math

def tri(n):
    return n*(n+1)/2

def is_square(integer):
    root = math.sqrt(integer)
    if int(root + 0.5) ** 2 == integer:
        return True
        return False

def next_tri(n, m):
    z = False
    i = 0
    while ((not z) and (i < m)):
        i += 1
        if is_square(tri(n)*tri(n+i)):
            return n+i
    return 0

def answers(n, m):
    list = [0]*n
    for i in range(1, n+1):
        list[i-1] = next_tri(i, m)
    return list

The first function here, tri, outputs triangular numbers; the second tells you if its input is a square, and is taken from this Stackoverflow answer. next_tri is where the “magic” happens; it outputs the index of the first triangular number k greater than n such that tri(n)*tri(k) is a square. But since this loop may, a priori, be infinite, we cut this off at k = n+m. If none of tri(n)tri(n+1), tri(n)tri(n+2), …, tri(n)tri(n+m) are square, it outputs 0. Finally, calling answers(n,m) outputs the list answers(1,m), answers(2,m), …, answers(n,m).

If you call answers(10, 1000) you get the output

[8, 24, 48, 80, 120, 168, 224, 49, 360, 440].

So, for example, tri(1) tri(8) = (1)(36) = 36 is square, namely 62; tri(2) tri(24) = (3)(300) = 900 = 302; and so on. If you ignore the eighth element of this list and stare for a moment, you start to see a quadratic! Namely,

tri(n) tri(4n(n+1))

appears to be a square. And in fact we can write

tri(n) tri(4n(n+1)) = {n(n+1) \over 2} {(4n^2+4n)(4n^2+4n+1) \over 2} = {n(n+1) \times 4(n(n+1))(2n+1)^2 \over 4} = [n(n+1)(2n+1)]^2.

This is an identity that’s not hard to prove, and answers Tanton’s question in the affirmative, but I think it’s hard to discover without computation. Of course you could do the computation by hand, but it’s 2012.

But this identity tells us that tri(8) tri(288) = (8)(9)(17)^2 — and this is true, but 288 is not the smallest solution to the problem that Tanton originally posed. In fact the computer search gives tri(8) tri(49) = 210^2. When do smaller solutions exist?


2 thoughts on “Square numbers as products of triangular numbers

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