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I'm interested in ${\bf integer}$ solutions of

$$abcd+1=(ecd-c-d)(fab-a-b)$$

subject to ${\bf a,b,c,d \geq 2}$, and ${\bf e,f \geq 1}$.

${\bf Questions:}$ Are there finitely many solutions? If no, is there a nice infinite family of solutions?

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  • $\begingroup$ can you prove they aren't all odd ? not all 1 mod 3 ? etc $\endgroup$ – Roddy MacPhee Apr 22 at 14:11
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    $\begingroup$ you can actually show they can't be all odd by expanding the RHS, subtraction leaving 1 on the LHS, and partial regrouping on the RHS. then applying parity arguments. $\endgroup$ – Roddy MacPhee Apr 22 at 14:22
  • $\begingroup$ @Roddy: sorry, I misread your comment. Yes, indeed at least one is even. $\endgroup$ – Jeff Apr 22 at 14:28
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    $\begingroup$ and at least one has a distinct remainder on division by 3. $\endgroup$ – Roddy MacPhee Apr 22 at 14:35
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$1=(f-1)abcd-fabc-fabd-acd+ac+ad-bcd+bc+bd$

Only if an odd number of odd terms exists, can this work. That takes at least one, even variable (not just term).

mod 3, we have that when turned into addition of possible negative equivalents, the 2 mod 3's either need to win by a number of terms that are 2 mod 3, or the 1 mod 3 terms need to win by a number of terms by a number of terms thst is 1 mod 3. 0 mod 3 has no effect, except forcing constraints on distribution of remaining terms. as 9 is not 1 or 2 mod 3 at least one term needs to be 0 mod 3. the remaining terms can split 3-5 for a 2 mod 3 win, or 7 terms can be non-zero mod 3 and get 4-3 split for a 1 mod 3 win. 6 non-zero terms have a 2-4, 2 mod 3 win. 5 has a 3-2, 1 mod 3 win. 4 has a 2 mod 3 win of 1-3. 3 has a 1 mod 3 win of 2-1. 2 has a 0-2 win for 2 mod 3. 1 has a 1-0 win for 1 mod 3.

you can do the same analysis mod 4 etc. then use CRT to combine them.

Edit 9 terms actually has a 2-7 split .

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There are finitely many such solutions.

${\bf Proof.}$

It follows directly from the following (not too hard to verify) statements:

(1) $e$ or $f$ is $1$. Once $f$ is fixed to be $1$, $e \leq 7$.

(2) At least one number from the set $S=\{a,b,c,d \}$ is $\leq 5$.

(3) For a fixed number from the set $S$, say $a$, there are at most finitely many triples $(b,c,d)$ satisfying this equation (for a proof see this question).

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