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It is known that the abc conjecture can't fail with polynomial identities.

Is the following special case of abc known?

Let $a,b,c,f$ be polynomials with integer coefficients satisfying $a+b=c+f$. Assume $f=0$ has infinitely many integers solutions $P_i$.

Then we have the integer abc triples $a(P_i)+b(P_i)=c(P_i)$ (after clearing the gcd).

abc implies that there are finitely many integer triples of quality $1+\epsilon$ for all fixed $\epsilon > 0$.

There is example with $f$ of degree $3$, which in general is elliptic curve, but in this case $f$ is singular and all integer triples for $f=0$ collapse to only one triple.

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    $\begingroup$ If $f$ is homogeneous, we expect a variety $f=0$ with infinitely many rational points to be contain either a rational curve or an abelian variety. In the rational curve case, we can make $a,b,c$ functions on this rational curve and do the usual Mason-Stothers argument. In the case of an elliptic curve this works as well, so I imagine for an abelian variety it does too. $\endgroup$
    – Will Sawin
    Commented May 12, 2019 at 13:21
  • $\begingroup$ @WillSawin What about this: $(3*x+2*y+1)^5-(1)^5 - (3*x + 2*y + 5) * (3*x + 2*y)^4 - (5) * (3*x + 2*y) * (18*x^2 + 24*x*y + 8*y^2 + 6*x + 4*y + 1) = 0$. If the cubic $(5) * (3*x + 2*y) * (18*x^2$ were elliptic curve wouldn't this violate abc? $\endgroup$
    – joro
    Commented May 12, 2019 at 15:40
  • $\begingroup$ But it's not an elliptic curve, so what's the problem? $\endgroup$
    – Will Sawin
    Commented May 12, 2019 at 16:23
  • $\begingroup$ I feel like your example would be clearer if you change variables $y'= 3x+2y$, as $3x+2y$ is repeated a lot in your expression. I get $(y'+1)^5 - 1^5 - (y'+ 5) (y'^4) = 5 y' (2 y'^2 + 2y'^2 + 1)$. This is not very relevant to abc as there are only finitely many solutions for $y'$. $\endgroup$
    – Will Sawin
    Commented May 12, 2019 at 16:25
  • $\begingroup$ @WillSawin Thanks. If you take $a,b,c$ with small radicals and set $f=a+b-c$ will this give constraints on the integral points of $f$ compared to randomly chosen $f$? $\endgroup$
    – joro
    Commented May 13, 2019 at 15:51

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