Conjecture on the square root of the sum of the squares of the prime factors of a number

Question

Let $A_{n}$ denote the square root of the sum of the squares of the prime factors of $n$.

For example, $A_{60}=\sqrt{2^2+2^2+3^2+5^2}\approx6.48$.

I have recently made the following observations:

• There does not exist $n$ such that $A_{n},A_{n+1},A_{n+2},A_{n+3}$ are all integers
• There does not exist $n$ such that $A_{n},A_{n+1},A_{n+2}$ are all non-prime integers

I tested these statements up to 1.2 billion, and both of them seemed to withstand.

I am basically trying to find out if either one of them has already been conjectured, proved or refuted.

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I have previously posted this question on Mathematics.

The original (and much longer) version can be found here.

My initial motivation was to depict every natural number as an $n$-dimensional rectangle with measures given by its prime-factorization, and then observe the ones which yield diagonal of an integer length.

Soon thereafter, it came clear to me that sequences of consecutive such numbers were rather sporadic. An example of a sequence of $3$ consecutive such numbers which yield diagonal of an integer length:

• $A_{2729}=\sqrt{2729^2}=2729$
• $A_{2730}=\sqrt{2^2+3^2+5^2+7^2+13^2}=16$
• $A_{2731}=\sqrt{2731^2}=2731$

The only answer I received gave a probabilistic argument to the fact that both conjectures are (probably) false, which I more or less understand, but I would nevertheless like to obtain a more absolute resolution.

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UPDATE:

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:

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

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UPDATE #$2$:

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

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Answer 1

It may be of interest to consider in general when A=A_n is integral. I will assume n is given and drop the subscript.

A is integral when n =p^k, for p prime and k a square.

A integral and n composite means n has (counting multiplicity) at least 4 prime factors.

Let m be a positive integer. There are integers a and b with b one of 1,3 , 9 such that mb2^a =n and A is integral. b could also be a prime or the square of a prime. It is possible that 2^a may be replaceable by p^a but I haven't thought much about it.

In all, it seems the answer given in the other thread has a plausible argument for both conjectures being negative, and the setup guarantees few if any small counterexamples. Finding a sequence of 5 composite consecutive n giving integral A seems natural; such n having as few as four prime factors in addition seems highly unlikely.