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Is there non-zero integer $N$ such that

$$ a^4+b^4-c^4=N \qquad (1)$$

has infinitely many integer solutions $(a,b,c)$ with $a,b \ne \pm c$?

(1) is a surface, so possible approach is to find genus 0 curve on it with infinitely many integral points.

Partial argument for positive answer is similar equation is possible over larger ring: $a^4+b^4+c^4=18$ over $\mathbb{Z}[i],i^2=-1$.

Let $a= x - y - 1 ,b= 2*y + 2 ,c= -x - y - 1$. Then $a^4+b^4+c^4=18 + q_1(x,y) q_2(x,y)$ where $q_1=x^2 + 3*y^2 + 6*y$. When $q_1=0$ we have solution and this is possible infinitely often over $\mathbb{Z}[i]$.

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    $\begingroup$ Not necessarily. First, there are a few obvious congruence obstructions, such as $N \equiv 3 \mod{8}$. More to the point, it can be shown that a zero density of integers $N$ are thus represented. $\endgroup$
    – Vesselin Dimitrov
    Commented Jan 3, 2018 at 14:49
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    $\begingroup$ Zero density seems like a hard theorem because conceivably a positive fraction of $N$ could allow solutions with dramatic cancellation (though we expect the number of representable $x < N$ to grow only as $x^{3/4 + o(1)}$). $\endgroup$
    – Noam D. Elkies
    Commented Jan 3, 2018 at 15:10
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    $\begingroup$ @VesselinDimitrov I am asking for existence of single N with infinitely many solutions. $\endgroup$
    – joro
    Commented Jan 3, 2018 at 15:39
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    $\begingroup$ @joro: Thanks for your edit! This is now a different question. I would a guess a negative answer. Incidentally you do not need to add the condition that $a,b \neq \pm c$. $\endgroup$
    – Vesselin Dimitrov
    Commented Jan 3, 2018 at 17:43
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    $\begingroup$ @joro: A similar question, but in the 'critical' rather than in your 'supercritical' case, was studied in a recent paper by Ghosh and Sarnak: arxiv.org/pdf/1706.06712.pdf (Integral points on Markoff type cubic surfaces). Its introduction raises exactly this kind of question. In the supercritical case, Vojta's conjecture prescribes that the solutions to $F(\mathbf{x}) = N$, for any $N$, are never Zariski-dense. Your question might be difficult to answer without assuming any such conjecture. $\endgroup$
    – Vesselin Dimitrov
    Commented Jan 3, 2018 at 22:44

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