# About primes of polynomial form.

It is well known that If a prime is of the form $x^2 + y^2$ where $x,y$ are positive integers , then there is only 1 such representation for that prime.

This uniqueness seems to be rare.

For instance the cube analog fails :

$$10^3 + 9^3 = 1^3 + 12^3 = 1729$$

Since $1729$ is a prime.

It Also fails for triangular numbers

$$31 = 28 + 3 = 21 + 10$$

Now I am aware of the connections to norms of a UFD. For instance $x^2 + y^2$ is the norm for the gaussian integers ( ring $Z(\sqrt -1)$ ).

But most integer-valued polynomials are not norms of rings.

So are there examples of this Uniqueness for primes of an integer-valued polynomial that is not the norm of a ring ?

Notice I do not accept $u^4 + v^4$ as an answer because that is Also of the form $x^2 + y^2$.

I would like to point out that polynomials with non-negative rational coëfficients are usually not norms ! In fact they are not When the Number of variables or the degree is above $2$.

I wonder If we have uniqueness for every prime $p$ of the form

$$p = a^6 + b^6 + c^6$$

Everytime I think or read about number-theory or ring theory this idea bothers me !

Appendix

Integer-valued polynomial means a polynomial with rational coëfficiënts that maps every integer to an integer. Example triangular numbers from $D(D+1)/2$.

• Do you mean homogeneous polynomials - otherwise eg primes of the form $n^4+5$ would qualify. – Mark Bennet Dec 29 '17 at 23:16
• I don't see how uniqueness of the representation is meaningful. For positive definite quadratic forms see Smith–Minkowski–Siegel_mass_formula. I'd say expressing the polynomial as a linear combination of (twisted) norms is the only way to obtain multiplicative identities. – reuns Dec 29 '17 at 23:28
• If $d$ is large enough (maybe even $d=5$) heuristics suggest that representations as $n = x^d+y^d$ are unique (not only when $n$ is prime, but for any $n$). – Rodrigo Dec 29 '17 at 23:32
• @Mark Well that is of very low degree (1). Second it has not been proven there are infinitely many of such. You have a point but ofcourse I am more into multivariable polynomials. – mick Dec 29 '17 at 23:35
• @Rodrigo that is similar to my sum of 6th powers. Im into things that are proven rather than heuristics ... – mick Dec 29 '17 at 23:38