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Can we find four natural numbers $X$, $Y$, $Z$, $T$, greater than 1 and different from each other, so that adding 1 to the product of any three of them, the sum is divisible by the fourth?

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    $\begingroup$ Is there a reason for the mismatch between question title and content, and for the metric-geometry and classical-analysis-and-odes tags (and probably the algebraic-geometry and prime-numbers tags as well)? $\endgroup$
    – user44191
    Commented Sep 1, 2023 at 18:35
  • $\begingroup$ True, but the question is interesting even so $\endgroup$ Commented Sep 1, 2023 at 18:42
  • $\begingroup$ @GHfromMO I may be missing something, but I don't see how that satisfies the question; $XYZ + 1 = 232$, which is not divisible by $T = 15$. $\endgroup$
    – user44191
    Commented Sep 1, 2023 at 19:05
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    $\begingroup$ It might be possible to find all such quadruples. $XYZT \mid XYZ + XYT + XZT + YZT + 1$ and number $1/X + 1/Y + 1/Z + 1/T + 1/(XYZT)$ is a positive integer. $\endgroup$ Commented Sep 1, 2023 at 20:01
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    $\begingroup$ @DenisShatrov That does limit it pretty heavily; the sum of five distinct integer reciprocals can only be an integer if either a) one of them is $1$, or b) the sum is exactly $1$, as $H_6 = \frac{59}{20} < 3$. $\endgroup$
    – user44191
    Commented Sep 1, 2023 at 20:29

1 Answer 1

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For four numbers, $2$, $3$, $7$ and $43$ do the job. In general, if $n$ is the amount of numbers and the product of all but one of them, say $T$, must be $-1$ modulo $T$, we can state the following:

  1. All numbers must be coprime to each other. For if not, take one of a pair that is not coprime. The product of the rest is not $-1$ modulo that number because then it would be coprime to that number.
  2. There is a combination of $n+1$ numbers with the same property by adding the product of all plus one. Since any product of $n$ of the combination containing this new number is still $-1$ modulo the missing number.
  3. Therefore each instance of such a combination leads to an infinite chain of such combinations.

Starting with $2$ is obvious since it equals the empty product plus one.

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