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I have two rational points. They are denoted below in black and they define a line segment. Find integer points on this segment.

a = {45/2, 0}; b = {45/3, -45/3};
Plot[2*x - 45, {x, 0, 23}, GridLines -> {Range[23], -Range[30]}, 
 AspectRatio -> 1, PlotRange -> {{0, 23}, {-23, 0}}, 
 Epilog -> {PointSize[Large], Point[{a, b}], Red, PointSize[Medium],
   Point[Table[{i, 2*i - 45}, {i, 15, 22}]]
   }]

It happens that I know the solutions, and they are denoted as red dots. The question is how to get them programmatically, not only in this but in all cases, for instance a = {21, 0}; b = {21, 10/5}.

line segment on a grid

Notice that unlike this post, i) the given points are rational, and ii) all solutions are required, not just positive.

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    $\begingroup$ I don't see where the problem is ... FindInstance, Reduce and Solve all work as expected: Solve[2*x - 45 == y && x >= b[[1]] && x <= a[[1]], {x, y}, Integers]. $\endgroup$
    – Domen
    Commented Aug 27 at 9:46
  • $\begingroup$ What do you mean by modifying the solutions? I posted a screenshot of the result as given by Mathematica (14.1). $\endgroup$
    – Domen
    Commented Aug 27 at 9:57
  • $\begingroup$ Obviously the conditions were x > 0 && y > 0 since the OP wanted to find coordinates (which) are positive integer. You don't have this condition, so you have to restrict the coordinates to your bounds. $\endgroup$
    – Domen
    Commented Aug 27 at 10:03

1 Answer 1

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Following the answer by Yves Klett on a practically identical question, this is easily computed via Solve (or any related equation-solving functions), you only have to be careful when you have a vertical line (ie. $x_1 = x_2$):

findIntegerSolutions[{x1_, y1_}, {x2_, y2_}] := Module[{line, x, y},
  line = If[x1 == x2, x == x1, y - y1 == (y2 - y1)/(x2 - x1) (x - x1)];
  SolveValues[line && {x} ∈ Interval[{x1, x2}] && {y} ∈ Interval[{y1, y2}], 
   {x, y}, Integers]
  ]

findIntegerSolutions[{45/2, 0}, {45/3, -45/3}]
(* {{15, -15}, {16, -13}, {17, -11}, {18, -9}, {19, -7}, {20, -5}, 
    {21, -3}, {22, -1}} *)

findIntegerSolutions[{21, 0}, {21, 10/5}]
(* {{21, 0}, {21, 1}, {21, 2}} *)

Or an even simpler solution, as given by @ydd:

findIntegerSolutions[a_, b_] := 
   Module[{x}, SolveValues[x ∈ Line[{a, b}], x, Integers]]
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  • $\begingroup$ Am I misunderstanding something or is just findIntegerSolutions[a_, b_] := SolveValues[x ∈ Line[{a, b}], x, Integers] not sufficient? $\endgroup$
    – ydd
    Commented Aug 27 at 21:27
  • $\begingroup$ @ydd, you're not missing anything :) You can probably add this also as an answer to the original post. $\endgroup$
    – Domen
    Commented Aug 28 at 6:52
  • $\begingroup$ oh cool. I'll credit you for the x localization because I probably should've included that. $\endgroup$
    – ydd
    Commented Aug 28 at 14:07

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