Hot answers tagged diophantine-equations
23
There is an especially useful function for this kind of task: FrobeniusSolve[{a, b, c}, d] for finding the list of all solutions to the equation a x + b y + c z == d, where a,b,c are given positive integers and d is an integer, while x,y,z are non-negative integers to be found. There are many solutions (884 of them):
FrobeniusSolve[{2, 3, 1}, 100] // Short
...
23
If not assumed otherwise m and n can be whatever, so you can do e.g. this :
Solve[ Prime[n] + Prime[m] == 100, {n, m}, Integers]
{{n -> 2, m -> 25}, {n -> 5, m -> 24}, {n -> 7, m -> 23}, {n -> 10, m -> 20},
{n -> 13, m -> 17}, {n -> 15, m -> 16}, {n -> 16, m -> 15}, {n -> 17, m -> 13},
{n -> 20, m ...
19
There are many ways to proceed, the best one uses FrobeniusSolve :
I
Since we know, that
a x + b == y /. Solve[{-4 a + b == 11, 16 a + b == -1}, {a, b}] // Simplify
{3 x + 5 y == 43}
we find
FrobeniusSolve[ {3, 5}, 43]
{{1, 8}, {6, 5}, {11, 2}}
a bit more straightforward way :
II
{x, y} /. Solve[ (a x + b == y /. Solve[ {-4 a + b == 11, 16 ...
14
The problem we encounter here is an instance of rather unexpected limitations of equation solving functionality (i.e. Modulus option in Reduce), e.g. this question : Strange behaviour of Reduce for Mod[x,1] provides another example which has been fixed in the newest version (9.0) of Mathematica. Since Modulus unexpectedly doesn't work here we can take ...
13
Is this what you are searching for?
a = {-4, 11};
b = {16, -1};
dy = (b[[2]] - a[[2]])/(b[[1]] - a[[1]]);
offset = u /. Solve[a[[2]] == dy*a[[1]] + u, u][[1]];
coords = {x,
y} /. {Reduce[y == dy*x + offset && x > 0 && y > 0, {x, y},
Integers] // ToRules}
(* {{1, 8}, {6, 5}, {11, 2}} *)
Graphics[{PointSize[Large], ...
12
A geometrical view of the solutions:
s = Solve[(x - 1)^2 + (y - 1)^2 + (z - 1)^2 == 49, {x, y, z}, Integers];
pts = {x, y, z} /. s;
subs = Subsets[pts, {2}];
minds = Union[dists = N[EuclideanDistance @@@ subs]][[1 ;; 3]];
Show[Graphics3D[Sphere[{1, 1, 1}, 13/2]],
Graphics3D[Line /@ Extract[subs, Position[dists, Alternatives @@ minds]]],
...
11
You can also use InterpolatingPolynomial with Solve, Reduce or Eliminate:
a = {-4, 11}; b = {16, -1};
coords = Solve[y == InterpolatingPolynomial[{a, b}, x] && 0 <= x <= 16&&0<=y,
{x, y}, Integers][[All, All, 2]];
(* or *)
coords={ToRules[Reduce[ y == InterpolatingPolynomial[{a, b}, x] &&
0 <= x <= ...
9
First, one should mathematically analyze the problem. Obviously there are infinitely many solutions of the form {1, y} and {x, 1}, as well as {x,y} where x == y. So we can exclude such solutions from our search.
Another point is remembering SystemOptions["ReduceOptions"]. There were questions dealing with them, so I'm not going to discuss these issues here; ...
8
Try :
Solve[(x - 1)^2 + (y - 1)^2 + (z - 1)^2 == 49, {x, y, z}, Integers]
or
Reduce[(x - 1)^2 + (y - 1)^2 + (z - 1)^2 == 49, {x, y, z}, Integers]
You can add inequalities as well as :
Solve[{(x - 1)^2 + (y - 1)^2 + (z - 1)^2 == 49, x != 1, y != 1, z != 1}, {x, y, z}, Integers]
8
Another possibility is perhaps FrobeniusSolve
For example,
Sort@Select[
ArrayFlatten[FrobeniusSolve[{1, 2, 3}, #] & /@ Range[2, 18, 2], 1],
FreeQ[#, 0] && #[[1]] < 10 && #[[2]] < 10 && #[[3]] < 10 &]
gives the following list (53 elements), essentially the same solution as the one posted by Daniel Lichtblau.
...
7
No reason not to have n a variable as well.
In[786]:= {x, y, z,
n} /. {ToRules[
Reduce[{(x + 2 y + 3 z) == 2 n, 1 <= x <= 9, 1 <= y <= 9,
1 <= z <= 9, 1 <= n <= 9}, {x, y, z, n}, Integers]]}
Out[786]= {{1, 1, 1, 3}, {1, 1, 3, 6}, {1, 1, 5, 9}, {1, 2, 1, 4}, {1,
2, 3, 7}, {1, 3, 1, 5}, {1, 3, 3, 8}, {1, 4, 1, 6}, {1, ...
7
FindInstance[{(x + 2 y + 3 z)/2 == 4 && 1<= x <= 9 && 1<=
y <= 9 && 1<= z <= 9}, {x, y, z}, Integers, 10]
gives two solutions: {{x -> 3, y -> 1, z -> 1}, {x -> 1, y -> 2, z -> 1}}
With relaxed constraints,
FindInstance[{(x + 2 y + 3 z)/2 == 4 && -100 <= x <= 100 && -100 ...
6
The Backsubstitution option will help here.
Reduce[
1 + x + y + x y + z + x z + y z - x y z == 0 && x >= y >= z >= 1, {x,
y, z}, Integers, Backsubstitution -> True]
(* (x == 5 && y == 4 && z == 3) || (x == 7 && y == 6 &&
z == 2) || (x == 8 && y == 3 && z == 3) || (x == 9 && ...
5
Sometimes tt is more reasonable to use Reduce :
Reduce[(x + 2 y + 3 z)/2 == 4 && 1 <= x <= 9 && 1 <= y <= 9 && 1 <= z <= 9, {x, y, z}, Integers]
(x == 1 && y == 2 && z == 1) || (x == 3 && y == 1 && z == 1)
For a larger domain
sols = Reduce[(x + 2 y + 3 z)/2 == 4 && -100 ...
4
The difference between $2^n$ and $n^2$ is that $2^n$ is not a function $\bmod 10$ -- that is, $2^{n+10}$ is not congruent to $2^n\bmod 10$. Further $2^n$ is only eventually periodic $\bmod 10^k$, $k \geq 2$. For instance $2^1$ is not congruent to any other $2^n \bmod 100$. On the other hand, polynomial functions are all functions $\bmod\, m$ : f[n+m] is ...
3
This isn't really an answer, but it's too big for a comment.
Here's some code to brute force solve the problem (since there wasn't any code provided in the question). It lists all the possible numbers; finds the ones whose product matches; then of those, finds the ones whose sum of squares match, etc..
Module[{h = 50, r, product, squared, fourth, sixth},
...
3
Artes's solution is the best, I think. If you just want to treat this as an ordinary Diophantine problem, you can do that with Solve[] (making this approach more or less equivalent to Yves's):
{p, q} = {-4, 11};
{r, s} = {16, -1};
{x, y} /. Solve[{(q - s) x - (p - r) y == -Det[{{p, q}, {r, s}}],
x > 0, y > 0, Min[p, r] < x < ...
3
There are also power series methods for counting these.
SeriesCoefficient[
x^(1 + 2 + 3)/(1 - x^1)*1/(1 - x^2)*1/(1 - x^3), {x, 0, 100}]
(* Out[118]= 784 *)
See also "Supplement to 'Perplexities Related to Fourier's 17 Line Problem'."
3
It seems to me that for your first stated problem there is a much better method than Solve or Reduce:
{m, k, p} = {16, 3, 6};
IntegerPartitions[m, {k}, Range@p]
{{6, 6, 4}, {6, 5, 5}}
If you want all permutations just use Permutations:
Join @@ Permutations /@ %
{{6, 6, 4}, {6, 4, 6}, {4, 6, 6}, {6, 5, 5}, {5, 6, 5}, {5, 5, 6}}
For your second ...
2
Try to read Documentation on functions Solve and Reduce and tutorial Solving Equations. Look through this forum, I bet this is a duplicate question. There is also a guide: Diophantine Equations.
{x1, x2, x3} /.
Solve[x1 + x2 + x3 == 16 && 1 <= x1 <= 6 && 1 <= x2 <= 6 &&
1 <= x3 <= 6, {x1, x2, x3}, Integers]
...
2
Cannot test this in Mathematica right now, but pattern matching should work to get the bits that appear next to x. So for instance you could use
Cases[Ans,(x == z_)->(z/.C[1]->0),Infinity]
That will capture all the terms where x appears on the left of == and will give return what on the right side of that equality with C[1] replaced by 0.
1
Suppose we know the equation of line through the two points, one can generate all points on the line with integer x and of them keep those with integer y. Without invoking solving function.
With[{x1 = -4, y1 = 11, x2 = 16, y2 = -1},
Table[{x, (y2 - y1)/(x2 - x1) (x - x1) + y1},
{x, x1, x2}]] // Cases[#, {_, _Integer}] &
(* {{-4, 11}, {1, 8}, {6, ...
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