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Is equation $$ (x+1)y^2-xz^2=x^3+2x+2 $$ solvable in integers?

Motivation: For a polynomial $P$ consisting of $k$ monomials of degrees $d_1,\dots,d_k$ and integer coefficients $a_1,\dots,a_k$, define its "size" as $H(P)=\sum_{i=1}^k |a_i|2^{d_i}$. If we order all equations by $H$, then the stated equation has $H=34$, and (after this previous equation has been solved) is currently the smallest cubic equation with open solvability problem.

Bogdan Grechuk
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  • Can rewrite as $(x+1)(y^2-2) = x(x^2+z^2)$. So $x+1|x^2+z^2$ and $x|y^2 -2$. From this it follows that $x+1$ has no 3 mod 4 prime factors, and so $x$ has at most one power of 2 in factorization. From $x|y^2-2$ aside from possibly a single 2, all prime factors of $x$ are 1 or -1 mod 8. – JoshuaZ Dec 04 '23 at 14:16
  • is there any simple reason why this is not solvable for $x=-14$ or other numbers of the form $-2(8t+7), t\geqslant 0$? – Fedor Petrov Dec 04 '23 at 15:58
  • @Fedor Petrov - this is the problematic case in my attempt as well! – Bogdan Grechuk Dec 04 '23 at 17:34
  • Equivalently, $u^2 - (x^2 + x)v^2 = x^3 + 2x + 2$ where $u = (x + 1)y + xz$, $v = y + z$. – Denis Shatrov Dec 04 '23 at 20:50
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    @Zack, thank you – Will Jagy Dec 04 '23 at 20:58
  • @Fedor: following JoshuaZ's idea, write $x^2+z^2=(x+1)(x+k)$, so that $y^2-2=x(x+k)$, and using a similar observation, the only primes dividing $(x+k)$ are $1\pmod 8$ and possibly a single $2$. If $k$ is even, then $x$ is odd, else $y^2-2 \equiv 0 \pmod 4$; and if $k$ is odd, then $x$ is even, else $x^2+z^2\equiv 0 \pmod 4$ with $x$ odd. So $k+x$ is odd and $1 \pmod 8$. When $x$ is even, it is $2 \pmod 4$, which means $z^2=kx+k+x \equiv 3 \pmod 4$. Maybe someone can finish the case where $k$ is even. – Zack Wolske Dec 04 '23 at 21:53
  • @ZackWolske I used to think that all cases except $x=-2(8t+7)$ are solvable by playing with such arguments. But I still do not see how an argument solving $x=-14$ can look like. – Fedor Petrov Dec 04 '23 at 22:58
  • @Fedor: suppose $x=-14$, $z^2+(-14)^2=-13(k-14)$, and $-14(k-14)=y^2-2$. Then $k$ is odd and $k-14$ is $1$ mod $4$, because it divides a sum of two squares. But then $-13(k-14)\equiv 3 \pmod 4$. – Zack Wolske Dec 04 '23 at 23:17
  • I do not know how to use these arguments to solve the cases where $x$ is odd. – Zack Wolske Dec 04 '23 at 23:19
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    @ZackWolske $k-14$ is negative, thus it is 3 mod 4, not 1 mod 4 – Fedor Petrov Dec 05 '23 at 04:50
  • @Fedor, yes, I see – Zack Wolske Dec 05 '23 at 22:25

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