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Let $a,b,c$ be integers. Let $p$ be a given prime. How to find the number of solutions to $p = a^3+b^3+c^3 - 3abc$ ?

Another question is ; let $w$ be a positive integer. Let $f(w)$ be the number of primes of type prime = $a^3+b^3+c^3 - 3abc$ below $w$. How does the function $f(w)$ behave ? How fast does it grow ? Are those primes of type $A$ $mod$ $B$ for some integers $A$ and $B$ ? How to deal with this ?

Can this be solved without computing the class number ?

mick
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3 Answers3

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Andre's quadratic form $a^2 + b^2 + c^2 - bc - ca - ab$ is only positive semidefinite. It is 0 if $a=b=c.$ Meanwhile, $$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; (b-c)^2,$$ $$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; (c-a)^2,$$ $$a^2 + b^2 + c^2 - bc - ca - ab \; \geq \; \frac{3}{4} \; (a-b)^2.$$ So the quadratic form cannot be 1 unless all three letters are within 1 of each other. For example, we cannot have some negative and some positive. Indeed, all possible 1's are arrangements of $(n,n,n+1)$ or $(n,n+1,n+1).$ Taking $n$ as nonnegative, we see that we get the value of the cubic either $3n+1$ or $3n+2.$ So all primes not 3 are represented. Not sure about 3.

EEEEDDDIITTT: Now that I think of it, cubes are $0,1,-1 \pmod 9.$ So, in order to get $a^3 + b^3 + c^3 - 3 a b c \equiv 0 \pmod 3,$ our choices are (A) $a,b,c$ are all divisible by 3, or (B) one of them is divisible by 3, the other two cubes give cancelling $\pm 1\pmod 9.$ In either case, the cubic is divisible by 9. So the prime 3 is not represented.

Will Jagy
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Hint: Your expression factors as $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Edit: The above hint was given when the problem specified that $a$, $b$, and $c$ are non-negative. Then finishing was a couple of steps away. Removing that condition introduces many possibilities, for $a+b+c=1$ has non-trivial solutions.

André Nicolas
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  • I knew that. So ? – mick Feb 10 '13 at 21:23
  • @mick, so in almost all cases, both of these factors are greater than 1, making the product not a prime. –  Feb 10 '13 at 21:32
  • Hmm if we express $c$ in terms of $a$ and $b$ by $c=1-a-b$. That seems usefull. – mick Feb 10 '13 at 21:44
  • I removed the condition for positivity ! – mick Feb 10 '13 at 21:46
  • @mick, it's still similar. Even without the positivity condition, $a^2+b^2+c^2-ab-bc-ca$ is very unlikely to be 1. So either this factor is 1 and $a+b+c$ is a prime, or $a+b+c = 1$ and $a^2+b^2+c^2-ab-bc-ca$ is a prime. The first case forces one two of $a,b,c$ to be off by at most 1, and second case $a^2+b^2+c^2-ab-bc-ca$ is a quadratic form in terms of $a-b$ and $a-c$, so you can say something primes represented by such form. –  Feb 10 '13 at 21:51
  • It needs to be primes of the form $3(a^2+b^2+ab+a+b)+1$ or $3(a^2+b^2+ab-a-b)+1$ if $a+b+c$ is a unit. – mick Feb 10 '13 at 21:54
  • @Sanchez, all primes other than 3 are represented. Maybe 3 as well, I did only your first case. – Will Jagy Feb 10 '13 at 21:54
  • Nope, no 3. see my answer. – Will Jagy Feb 10 '13 at 22:02
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We may also prove that there are infinitely many primes when $a+b+c=1$. This answer along with Andre Nicolas' answer and Will Jagy's answer allows us to calculate the exact number of representations $a^3+b^3+c^3-abc=p$ for each prime $p$.

We begin with the factorization $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$ When $a+b+c=1$, we have that $$a^2+b^2+c^2-ab-bc-ca=3a^2+3ab+3b^2-3a-3b+1.$$ Rearranging, this becomes $$\frac{3}{4}\left(2a+b-1\right)^2+\frac{1}{4}\left(3b-1\right)^2.$$ If $x=a+2b-1,$ and $y=a-b,$ then the above is $$\frac{3}{4}\left(x+y\right)^{2}+\frac{1}{4}\left(x-y\right)^{2}=x^{2}+xy+y^{2}.$$ This is the norm form for $\mathbb{Z}\left[\zeta_{3}\right],$ the Eisenstein integers, and we know exactly which primes $p$ can be represented as $$p=x^{2}+xy+y^{2}.$$ To count the number of such representations, see my answer here. Note that this happens if and only if $p\equiv1 \pmod{6}$, and so we see that when $a+b+c=1$, the polynomial $a^3+b^3+c^3-3abc$ represents precisely those primes which are $1$ modulo $6$.

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Eric Naslund
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