# The Diophantine equation $(xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers with $z > a^2+2a$

Problem. Let $$a$$ be a positive integer that is not a perfect cube. Show that the Diophantine equation $$(xz+1)(yz+1)=az^{3}+1$$ has no solutions in positive integers $$x, y, z$$ with $$z > a^{2}+2a$$.

If this result is true, we can prove that for a given integer $$a$$, $$a \not= m^3$$, there are finitely many Fermat pseudoprimes of the form $$ap^{3}+1$$ to any base $$b>1$$ where $$p$$ runs through the primes.

Theorem 1. Let $$n = ap^{3}+1$$, $$a \not = m^3$$, $$p > a^2+2a$$ is prime. If there exists an integer $$b$$ such that $$b^{n-1} \equiv 1 \ ($$mod $$\ n)$$ and $$b^{a} \not\equiv 1 \ ($$mod $$\ n)$$ then $$n$$ is prime.

Proof. It can be shown that if $$p \ | \ \phi(n)$$ then $$n=(sp+1)(tp+1)$$ for some integer $$sp+1$$ and prime $$tp+1$$. Because the Diophantine equation $$(sp+1)(tp+1)=ap^3+1$$ has no solutions in positive integers $$s, t$$ when $$p > a^2 +2a$$, we must have $$s = 0$$. Therefore if $$p \ | \ \phi(n)$$, $$n = tp+1$$ is prime. Assume $$n$$ is composite, $$b^{n-1} \equiv 1 \ ($$mod $$\ n)$$ and $$b^{a} \not\equiv 1 \ ($$mod $$\ n)$$, we have $$ord_n b \ | \ \phi(n)$$. If $$p \ | \ ord_n b$$, then $$p \ | \ \phi(n)$$, a contradiction because $$n$$ is assumed composite therefore $$p \ \not | \ ord_n b$$. We also have $$ord_n b \ | \ n-1 = ap^3$$. Because $$p \ \not | \ ord_n b$$, we must have $$ord_n b \ | \ a$$. Hence $$b^a \ \equiv 1 \ (mod \ n)$$ which contradicts the hypothesis therefore $$n$$ must be prime.

Remark. For all $$n > b^a$$, we have $$b^a \ \not\equiv 1 \ (mod \ n)$$ therefore there are finitely many Fermat pseudoprimes of the form $$ap^3+1$$ for a given positive integer $$a$$ that is not a perfect cube.

A lengthy and incomplete solution to this problem can be found here https://math.stackexchange.com/questions/3842292/prove-that-the-diophantine-equation-xz1yz1-az3-1-has-no-solutions-in

• my solution was complete. The discussion split naturally across $a^{2/3}$ which was not an integer because $a$ itself was not a cube. The reason for doing it this way was computer modelling with relatively small values of the parameters, in which it was obvious that some quantity increased up until $a^{2/3}$ and then decreased past $a^{2/3}$ Computers don't prove things, but they may show patterns that can then be proved. Commented Mar 20, 2021 at 16:36
• Found your email of October 21. On October 22 I sent you a Latex pdf with a bit more detail. Again, if it does not work for you, I suggest you program the problem with small values of $a.$ I defined letters $w,t$ with $w^2 =x^2 y^2 + 4a (x+y)$ and $t = \frac{w-xy}{2}.$ Then proofs $t \leq a-1.$ Next, when $a^{2/3} < t \leq a-1$ we find $z < a^{4/3}.$ Third, when $2 \leq t <a^{2/3}$ we find $z < \frac{a^2+a}{2}.$ Fourth, when $t=1,$ we find $z \leq a^2+2a.$ Nothing easy or quick about this; if necessary, get help to program this Commented Mar 20, 2021 at 17:22
• I finally figured your proof out. Have made it concise below
– ASP
Commented Mar 29, 2021 at 10:46

The given equation $$(xz+1)(yz+1)=az^3+1$$ can be rewritten as $$az^2-xyz-(x+y)=0$$. We shall show that for any solution $$(x,y,z)$$, we have $$z \le a^2+2a. \$$ Note that $$z \ | \ x+y$$, therefore $$z \le x+y. \$$ Treating $$x, y$$ as constants, the only positive solution for $$z \$$ is $$$$z = \frac{xy+\sqrt{x^2y^2+4a(x+y)}} {2a}$$$$ In order for $$z$$ to be rational, the discriminant must be a perfect square. Therefore $$w^2 = x^2y^2+4a(x+y)$$. We see that $$w > xy$$ and $$w \equiv xy \ ($$mod$$\ 2)$$. We can write $$w = xy + 2t$$, $$t > 0$$. Substituting $$w$$ above, $$(xy+2t)^2 = x^2y^2+4a(x+y)$$. Expanding and simplifying, $$txy - ax - ay +t^2 = 0$$. Multiplying through by $$t$$ and factoring, $$(tx-a)(ty-a)=a^2 - t^3$$. We must have $$t \le a-1$$ otherwise $$RHS<0$$ and $$LHS \ge 0$$. Because $$a$$ is not a perfect cube, $$a^2 - t^3 \not = 0$$. The remainder of the proof utilizes the result: If $$ab = c \$$ where $$a,b, c \not = 0$$ are integers then $$a+b \le c+1$$ if $$c>0$$ and $$a+b \le -(c+1)$$ if $$c < 0$$. We now consider two cases:
Case $$1: \$$ $$a^2 - t^3 >0 \ ;$$ Using the result above on the factored equation, we have $$(tx-a)+(ty-a) \le a^2 - t^3+1 \le a^2$$. Hence, $$z \le x+y \le (a^2+2a)/t \le a^2+2a \\\\$$.
Case $$2: \$$ $$a^2 - t^3 < 0 \ ;$$ As in case $$1$$, we have $$(tx-a)+(ty-a) \le t^3 - a^2-1 \$$, $$x+y \le t^2 - (a^2-2a+1)/t < t^2$$. Hence , $$z \le x+y < t^2 \le (a-1)^2 < a^2 +2a$$
• good. Two typos in this , $t^2xy - tax - tay +t^2 = 0$. should say $t^2xy - tax - tay +t^3 = 0$. I see what you did now, in the last line you don't care about the stronger $a^{4/3}$ bound available for this case, you settle for the value at $t=a-1$ Commented Mar 29, 2021 at 16:43
• Fixed the typo. Yea I made use of the inequality $t \le a-1$