Prime factorizations that yield hyperrectangles with integer diagonals

Question

I have been looking into $n$-dimensional rectangles (aka hyperrectangles) with measures given by any orderless prime-factorization of a natural number, where the diagonal is of an integer length.

Let's denote these as perfect rectangles.

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Obviously, every prime number yields a perfect ($1$-dimensional) rectangle.

As a more generalized definition - any number whose prime-factorization is given by $\overbrace{{p}\times\dots\times{p}}^\text{N times}$ with $N$ being a square yields a perfect $N$-dimensional rectangle. For example:

$81$ yields a perfect $4$-dimensional rectangle with diagonal length $\sqrt{3^2+3^2+3^2+3^2}=6$.

However, there are other types of numbers that yield perfect rectangles. For example:

$48$ yields a perfect $5$-dimensional rectangle with diagonal length $\sqrt{2^2+2^2+2^2+2^2+3^2}=5$.

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I have been searching for sequences of consecutive numbers that yield perfect rectangles.

For example, $16$ and $17$ yield perfect rectangles with diagonal lengths $4$ and $17$ respectively.

$2729-2730-2731$ yield perfect rectangles with diagonal lengths $2729-16-2731$ respectively.

Searching up to $16$ million, I have not been able to find any sequence longer than $3$ numbers.

In addition, I have noticed that in every sequence of $3$ numbers, one or two of them are prime.

So my question is - has either one of these two statements been conjectured, proved or refuted?

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C-code is given below (should anyone wishes to extend the tested range):

#include <math.h>
#include <stdio.h>

#define RANGE 16000000
#define SEQUENCE_LEN 4

typedef unsigned char      uint08;
typedef unsigned int       uint32;
typedef unsigned long long uint64;

uint08 sieve[RANGE] = {0};
uint32 prime[RANGE] = {0};
uint32 numOfPrimes  =  0 ;

void CalcAuxiliaryData()
{
    uint32 i,j;

    uint32 root = (uint32)sqrt((double)RANGE);

    for (i=2; i<=root; i++)
    {
        if (sieve[i] == 0)
            for (j=i+i; j<RANGE; j+=i)
                sieve[j] = 1;
    }

    for (i=2; i<RANGE; i++)
    {
        if (sieve[i] == 0)
            prime[numOfPrimes++] = i;
    }
}

uint32 CalcDiagonalLen(uint32 n)
{
    uint32 i;

    uint64 square;
    uint32 length;

    if (sieve[n] == 0)
        return n;

    square = 0;
    for (i=0; i<numOfPrimes && n>1; i++)
    {
        uint32 p = prime[i];
        uint64 pp = (uint64)p*p;
        while (n%p == 0)
        {
            n /= p;
            square += pp;
        }
    }

    length = (uint32)sqrt((double)square);
    if ((uint64)length*length == square)
        return length;

    return 0;
}

int main()
{
    uint32 i;

    uint32 sequence_len;
    uint32 diagonal_len;

    CalcAuxiliaryData();

    sequence_len = 0;
    for (i=2; i<RANGE; i++)
    {
        diagonal_len = CalcDiagonalLen(i);
        if (diagonal_len == 0)
        {
            sequence_len = 0;
        }
        else
        {
            printf("%u %u\n",i,diagonal_len);
            if (++sequence_len == SEQUENCE_LEN)
                break;
        }
    }

    return 0;
}

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UPDATE:

Using this implementation advice, I have verified both statements up to $1.2$ billion.

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UPDATE #$2$:

Checking up to $2$ billion, I have counted $1585$ triplets:

• The trivial triplet $1-2-3$
• $2$ triplets of the form $C-C-P$
• $4$ triplets of the form $P-C-C$
• $7$ triplets of the form $C-P-C$
• $1571$ triplets of the form $P-C-P$

I have also encountered the following quadruplet, which refutes the first conjecture:

• $A_{1776463301}=\sqrt{1776463301^2}=1776463301$
• $A_{1776463302}=\sqrt{2^2+3^2+173^2+857^2+1997^2}=2180$
• $A_{1776463303}=\sqrt{1776463303^2}=1776463303$
• $A_{1776463304}=\sqrt{2^2+2^2+2^2+7^2+11^2+179^2+16111^2}=16112$

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UPDATE #$3$:

Up to $4$ billion, I have counted $28$ pairs of consecutive non-primes which yield perfect rectangles, but I have not encountered a single triplet of consecutive non-primes which yield perfect rectangles.

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Answer 1

I would conjecture quite the opposite: for any $k$ there should be infinitely many integers $n$ such that $n,n+1,n+k-2$ all yield perfect rectangles. Indeed, these examples can avoid primes altogether.

Given a large number $x$, let's consider all primes less than $x^{1/k}$; there are approximately $k x^{1/k}/\log x$ such primes by the prime number theorem. Now let's consider all numbers that are the product of exactly $k$ distinct primes less than $x^{1/k}$; there are approximately $\frac1{k!} ( k x^{1/k}/\log x) ^k \approx C_1(k) x/(\log x)^k$ such numbers, and any such number is less than $x$. (The leading constant, depending on $k$, won't matter much.)

For any number of this form, the sum of the squares of the prime factors is at most $kx^{2/k}$. Assuming the heuristic that these sums of squares are distributed randomly, the "probability" that the sum of the squares of the prime factors is a perfect square is at least $\frac1{\sqrt k} x^{-1/k}$. Therefore, there should be about $C_2(k) x^{1-1/k}/(\log x)^k$ numbers, consisting of $k$ small primes multiplied together, that yield perfect rectangles. In other words, the probability that a randomly chosen number from $[1,x]$ has this property is about $C_2(k) x^{-1/k}/(\log x)^k$.

If we further assume the heuristic that these numbers are randomly distributed in the interval $[1,x]$, then the probability of any given $n,n+1,\dots,n+k-2$ consisting entierly of integers of this form is about $C_3(k) x^{-(k-1)/k}/(\log x)^{k(k-1)}$. Summing this probability over $1\le n\le x$, we discover that the expected number of sets of $2k-1$ consecutive numbers with this property is about $C_4(k) x^{1/k}/(\log x)^{k(k-1)}$, which in particular goes to infinity with $x$.

(Some slight modifications of this argument need to be made: for example, every 2nd number in the sequence must be a multiple of 2, every 3rd must be a multiple of 3, and so on for all primes less than $k$. I believe one can incorporate this modification with no substantive change to the overall heuristic.)