$(x^3+1)\cdot (y\cdot z^2+1)=43\cdot w$

Question

Consider the diophantine equation:

$(x+1)\cdot (x^2-x+1)\cdot (y\cdot z^2+1)=43\cdot w$ with x,y,z,w>0.

A set of solutions should be given if we set x=7 because in this case the factor $43$ emerges from $7^2-7+1=43$

But what about the other factors?

In particular are there infinitely many solutions to this equation?

Second question:

suppose instead that x,z,y are three consecutive primes.

Can it be proven that $x=7$ $z=11$ and $y=13$ is the only triple of x y z which lead to a solution?

Answer 1

Claim: There are infinite solutions to the equation consisting of three consecutive primes $(x,y,z)$

Proof: Note that the equation has a solution $(x,y,z)$ for $x,y,z$ primes if $43 \mid x^3+1$. This occurs if $43 \mid x^2-x+1$. We can see that this holds if $x \equiv 7 \mod 43 \iff x=43a + 7$, given Dirichlet's theorem we know that there are infinite primes of this form and we only have to take one of them and pick the next two consecutive primes for $y$ and $z$.

Examples: $(x,y,z) = (179,181,191), (523,541,547), (953,967,971) \ldots$

Remark: As Mike pointed out, note that $43$ could also divide $x+1$ (or $yz^2+1$) and we would still have positive solutions; nevertheless, we don't have to consider those cases when trying to prove the claim above.

Answer 2

Let $x$, $y$, $z$ and $w$ be positive integers such that $$(x+1)(x^2-x+1)(yz^2+1)=43w.\tag{1}$$ That is to say, the product on the left hand side is divisible by $43$. Because $43$ is a prime number this happens if and only if $43$ divides one of the three factors on the left hand side. We have \begin{eqnarray} 43\mid x+1\qquad&\Leftrightarrow&\qquad x\equiv42\pmod{43},\\ 43\mid x^2-x+1\qquad&\Leftrightarrow&\qquad x\equiv\phantom{4}7\pmod{43}\ &\vee&\ x\equiv\ \, 37\pmod{43},\\ 43\mid yz^2+1\qquad&\Leftrightarrow&\qquad z\not\equiv\phantom{4}0\pmod{43}\ &\wedge&\ y\equiv z^{-2}\pmod{43}. \end{eqnarray} This completely characterizes the positive integral solutions to $(1)$. In particular we see that if $x$ is a prime congruent to $7$, $37$ or $42$ modulo $43$, and $y$ and $z$ are the next two primes, we also get a solution to $(1)$. By Dirichlet's theorem on primes in arithmetic progressions, there are infinitely many such primes.