How to prove than $a+b+c = 2^n-1$ and $a^2+b^2+c^2 = (4^n-1)/3$ have integer solutions only with Mersenne exponent or exponents of Mersenne exponent?

Question

I noticed something with Mersenne numbers : you can write it with the form $a+b+c = 2^n-1$ and $a²+b²+c² = (4^n-1)/3$ when $n$ is a odd Mersenne exponent (3, 5, 7, 13 ...) or an exponent of a odd Mersenne exponent ($3^2, 5^4, 7^3 ...$)

For example with Mersenne exponent :

• $4+2+1 = 7 = 2^3-1$ and $4^2+2^2+1^2 = 21 = (4^3-1)/3$
• $14+9+8 = 31 = 2^5-1$ and $14^2+9^2+8^2 = 341 = (4^5-1)/3$
• $2760+2761+2670 = 8191 = 2^{13}-1$ and $2760^2+2761^2+2670^2 = 22369621 = (4^{13}-1)/3$

And exponent of Mersenne exponent :

• $44732914+44738572+44746241 = 134217727 = 2^{27}-1$ and $44732914^2+44738572^2+44746241^2 = 6004799503160661 = (4^{27}-1)/3$ with $27 = 3^3$

You can notice there are each time two even numbers and one odd number for $a, b$ and $c$.

Another observation : it seems there are no integer solution for $a, b, c$ when $2^n-1$ a composite Mersenne numbers like $2^{11}-1$. Apparently, you can't write $a+b+c = 2^{11}-1$ and $a^2+b^2+c^2 = (4^{11}-1)/3$ for this case.

Is there a way to explain that ? I don't know to start for proving it.

Answer 1

Only a partial answer for now. We can expect the unicity of the decomposition of $(4^n-1)/3$ as a sum of three squares up to order and signs. A formula due to Gauss (see the Wikipedia article about the sum of squares function) gives $r_{3}(n)=12h(-4n)$ when $n$ is a squarefree integer such that $n\pmod 8\not\in\{3,7\}$. A computation on wolfram alpha of the class number of $-84$ gives $4$ as a result, so $r_{3}(21)=12\times 4=3!\times 2^3$. Hence in the first case, the decomposition is unique.

Your question evocates the famous Fermat theorem proven by Euler saying $p=4m+1$ is prime iff $r_{2}(p)=8$, so I expect something along these lines: $2^{n}-1$ is prime iff there is a unique triplet of positive integers $(a,b,c)$ with $a<b<c$ such that $a^{k}+b^{k}+c^{k}=((2k)^{n}-1)/(2k-1)$ whenever $k\in\{1,2\}$.

If there are infinitely many îmaginary quadratic number fields of class number $4$, we can expect the existence of infinitely many Mersenne primes, and thus of even perfect numbers.

Edit: the class numbers corresponding to the second and third examples are $28$ and $3476$, so my unicity conjecture is false.