# A specific Diophantine equation restricted to prime values of variables.

Consider the following Diophantine equation $$x^2+x+1=(a^2+a+1)(b^2+b+1)(c^2+c+1).$$ Assume also that $x,a,b,c,a^2+a+1,b^2+b+1, c^2+c+1$ are all primes. We'll call such a quadruplet $(x,a,b,c)$ a triple threat.

I'd like to be able to show that there are no triple threats. I've been able to prove the following weaker result that if $(x,a,b,c)$ is a triple-threat and $x \equiv 1$ (mod 5), then none of $a,b$ or $c$ can be $1$ (mod 5). For my purposes, this is useful, but I can get a stronger result with a simpler proof if I can get that there are no triple threats.

Let me briefly sketch the idea of the proof: Most of what I have been able to do has been from looking at the more general equation

$$x^2+x+1=(a^2+a+1)(b^2+b+1)p$$ where $x,a,b,p,a^2+a+1$ and $b^2+b+1$ are all prime. Note that this equation does solutions (and probably has infinitely many). Call this an almost-triple threat.

The key approach is showing the following sort of result: If $a \leq b$, and we have some specific modulo restriction on $x,a,b,p$ then one must have $a^2+a+1<p$. If one then applies to the original triple threat equation one can in some circumstances get a contradiction by forcing that $a^2+1<c^2+c+1<b^2+b+1<a^2+a+1$.

Let me briefly sketch how in most contexts I've obtained that $a^2+a+1<p$.

My primary method of attack on this equation has been to rewrite it as $$(x-a)(x+a+1)=((b^2+b+1)p-1)(a^2+a+1)$$ and $$(x-b)(x+b+1)=((a^2+a+1)p-1)(b^2+b+1).$$

Since $a^2+a+1$ and $b^2+b+1$ are assumed to be prime, one has that either the pair of relations $a^2+a+1|x-a$ and $x+a+1|(b^2+b+1)p-1$, or the pair of relations $a^2+a+1|x+a+1$ and $x-a|(b^2+b+1)p-1$. One gets similar pairings using the second version above with $b^2+b+1$. One has then four cases:

I) $a^2+a+1|x-a$ and $b^2+b+1|x-b$

II) $a^2+a+1|x-a$ and $b^2+b+1|x+b+1$

III) $a^2+a+1|x+a+1$ and $b^2+b+1|x-b$.

IV) $a^2+a+1|x+a+1$ and $b^2+b+1|x+b+1$.

It is not too hard to show in Cases II and III that one always has $a^2+a+1<p$. The difficulty is primarily in Cases I and IV. In Case I one gets with a little algebra that $x+a+1|p(a+b+1) + \frac{x-b}{b^2+b+1}$. One then has $k(x+a+1)= p(a+b+1) + \frac{x-b}{b^2+b+1}$. For many choices of $x,a,b$ mod 5 one can force $k$ to be large. Say $k \geq 3$. Then one has $$a+x+1 \leq \frac{ p(a+b+1) + \frac{x-b}{b^2+b+1}}{3},$$ and one can get that $a^2+a+1<p$ from this inequality and a little work. (I can expand on this part if necessary.)

The main difficulty right now is dealing with Case IV. I can handle Case IV under some small modulus assumptions, but I cannot prove that $a^2+a+1<p$ in Case IV unconditionally.

In Case IV one has with a little work that $x-a|p(a+b+1)-\frac{x+b+1}{b^2+b+1}$ and $x-b|p(a+b+1)-\frac{x+a+1}{a^2+a+1}$ but it seems tough to use these to get a tight enough bound to conclude what is wanted without having some extra modulus restriction on the variables. Any thoughts on either clearing out Case IV, or any other method of showing that triple threats don't exist via some other method would be appreciated.

• You know that none of a,b,c,x can be 1 mod 6, right? Gerhard "Not Sure About One Mod Five" Paseman, 2018.08.13. – Gerhard Paseman Aug 13 '18 at 21:02
• @Gerhard Paseman Yes, aware of that. That and some other modulus restrictions which arise naturally from the problem are used in the proof of my current result on the matter. To get the sort of result I mentioned in Case I one does need to use mod 3.. – JoshuaZ Aug 13 '18 at 21:13
• So, you have seven numbers, and you want to know whether they can all be prime. Do you have examples where the equation holds and six of the seven numbers are prime? Have you done a computer search for triple threats, and, if so, how extensive? – Gerry Myerson Aug 13 '18 at 23:19
• I haven't done a systematic search for six of seven, but I did do a search for triple threats which didn't find any with x<10^6. That said, there are examples of solutions to the almost triple threats but I haven't looked for them systematically. Heuristically, I expect that for any primes a and b where a^2+a+1 and b^2+b+1 are both prime there should be infinitely many solutions to x^2+x+1=(a^2+a+1)(b^2+b+1)p with p prime. – JoshuaZ Aug 13 '18 at 23:28

Let $j$ be $a$, $b$, or $c$ in $x^2+x+1=(a^2+a+1)(b^2+b+1)(c^2+c+1)$. We have
$x^2+x+1 = 0\mod j^2+j+1$, $x^2+x+1+(j-x-1)(j^2+j+1) = 0\mod j^2+j+1$, $(x-j)(x-j^2) = 0\mod j^2+j+1$.
Since $j^2+j+1$ is prime, $x = j\mod j^2+j+1$ or $x = j^2\mod j^2+j+1$.
This gives 8 possible chinese remainders for $x\mod (a^2+a+1)(b^2+b+1)(c^2+c+1)$. These remainders are expected to have order of magnitude $(abc)^2$, whereas x is in the vicinity of $(a+1/2)(b+1/2)(c+1/2)$.