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December 10

More on the above conjecture

Above I posed:

Template:Serif

If true, it implies no natural prime is a prime in the ring $\mathbb{Z}[e^{\pi i/4}]$.

The absolute-value bars are not necessary. A number that can be written in the form $-(2a^2-b^2)$ is also expressible in the form $+(2a^2-b^2).$

It turns out (experimentally; no proof) that a number that can be written in two of these forms can also be written in the third form. The conjecture is not strongly related to the concept of primality, as can be seen in this reformulation:

Template:Serif

The first few numbers that cannot be written in any one of these three forms are

$15,$ $21,$ $30,$ $35,$ $39,$ $42,$ $55,$ $60,$ $69,$ $70,$ $77,$ $78,$ $84,$ $87,$ $91,$ $93,$ $95.$

They are indeed all composite, but why this should be so is a mystery to me. What do $2310=2\times 3\times 5\times 7\times 11,$ $5893=71\times 83$ and $7429=17\times 19\times 23,$ which appear later in the list, have in common? I see no pattern.

It seems furthermore that the primorials, starting with $5\mathrm{\#}=30,$ make the list. (Checked up to $37\mathrm{\#}=7420738134810.$)  --Lambiam 19:23, 10 December 2024 (UTC)

Quick note, for those like me who are curious how numbers of the form $-(2a^2-b^2)$ can be written into a form of $2a^2-b^2$, note that $2a^2-b^2=(2a+b{)}^2-2(a+b{)}^2$, and so $2a^2-b^2=-p\Rightarrow p=2(a+b{)}^2-(2a+b{)}^2$. GalacticShoe (talk) 02:20, 11 December 2024 (UTC)

A prime is expressible as the sum of two squares if and only if it is congruent to $1(\mathrm{mod}4)$, as per Fermat's theorem on sums of two squares. A prime is expressible of the form $2a^2+b^2$ if and only if it is congruent to $1,3(\mathrm{mod}8)$, as per OEIS:A002479. And a prime is expressible of the form $2a^2-b^2$ if and only if it is congruent to $1,7(\mathrm{mod}8)$, as per OEIS:A035251. Between these congruences, all primes are covered. GalacticShoe (talk) 05:59, 11 December 2024 (UTC)

More generally, a number is not expressible as:

1. $a^2+b^2$ if it has a prime factor congruent to $3(\mathrm{mod}4)$ that is raised to an odd power (equivalently, $3,7(\mathrm{mod}8)$.)
2. $2a^2+b^2$ if it has a prime factor congruent to $5,7(\mathrm{mod}8)$ that is raised to an odd power
3. $2a^2-b^2$ if it has a prime factor congruent to $3,5(\mathrm{mod}8)$ that is raised to an odd power

It is easy to see why expressibility as any two of these forms leads to the third form holding, and also we can see why it's difficult to see a pattern in numbers that are expressible in none of these forms, in particular we get somewhat-convoluted requirements on exponents of primes in the factorization satisfying congruences modulo 8. GalacticShoe (talk) 06:17, 11 December 2024 (UTC)

Thanks. Is any of this covered in some Wikipedia article?  --Lambiam 10:06, 11 December 2024 (UTC)

All primes? 2 is not covered! 176.0.133.82 (talk) 08:00, 17 December 2024 (UTC)

$2$ can be written in all three forms: $2=1^2+1^2=2\cdot 1^2+0^2=2\cdot 1^2-0^2.$  --Lambiam 09:38, 17 December 2024 (UTC)

I don't say it's not covered by the conjecture. I say it's not covered by the discussed classes of remainders. 176.0.133.82 (talk) 14:54, 17 December 2024 (UTC)

Odd prime, my bad. GalacticShoe (talk) 16:38, 17 December 2024 (UTC)

Assume p is 3 mod 4. Suppose that (2|p)=1. Then $x^4+1\equiv (x^2+\lambda x+1)(x^2-\lambda x+1)(\mathrm{mod}p)$ where ${\lambda }^2-2\equiv 0$. Because the cyclotomic ideal $(p,{\zeta }^2+\lambda \zeta +1)$ has norm $p^2$ and is stable under the Galois action $\zeta \mapsto 1/\zeta$ it is generated by a single element $a{\zeta }^2+b\zeta +a$, of norm $(2a^2-b^2{)}^2$.

If (2|p)=-1, then the relevant ideal is stable under $\zeta \mapsto -1/\zeta$ and so is generated by $a{\zeta }^2+b\zeta -a$, of norm $(2a^2+b^2{)}^2$. Tito Omburo (talk) 14:43, 11 December 2024 (UTC)