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Writing a program that will filter the integer solutions

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Writing a program that will filter the integer solutions

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2

$begingroup$

I've the following equation:

b==(-12 + 

  Sqrt[3] Sqrt[

   3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 

    4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

And $r$ is a given number and I'll run $a$ from $2$ to a given number n, which are all integers.

How can I write a code that will give me only an output of $b$ when the equation gives an integer back?

---

For example, when I use $r=4$ and use the following code:

Table[Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6], {a, 1, 100, 1}]

It will find that $a=24$ gives $b=70$, so I want to see only that solution.

Maybe I can use the If[] function?

equation-solving programming code-request system code

share|improve this question

asked 8 hours ago

JanJan

25022 silver badges1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  try Table[If[IntegerQ[Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6]],Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6],Nothing], {a, 1, 100, 1}] or shorter Table[If[IntegerQ[#,#,Nothing]&@(Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6]), {a, 1, 100, 1}]?
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

I've the following equation:

b==(-12 + 

  Sqrt[3] Sqrt[

   3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 

    4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

And $r$ is a given number and I'll run $a$ from $2$ to a given number n, which are all integers.

How can I write a code that will give me only an output of $b$ when the equation gives an integer back?

---

For example, when I use $r=4$ and use the following code:

Table[Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6], {a, 1, 100, 1}]

It will find that $a=24$ gives $b=70$, so I want to see only that solution.

Maybe I can use the If[] function?

equation-solving programming code-request system code

share|improve this question

asked 8 hours ago

JanJan

25022 silver badges1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  try Table[If[IntegerQ[Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6]],Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6],Nothing], {a, 1, 100, 1}] or shorter Table[If[IntegerQ[#,#,Nothing]&@(Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6]), {a, 1, 100, 1}]?
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

I've the following equation:

b==(-12 + 

  Sqrt[3] Sqrt[

   3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 

    4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

And $r$ is a given number and I'll run $a$ from $2$ to a given number n, which are all integers.

How can I write a code that will give me only an output of $b$ when the equation gives an integer back?

---

For example, when I use $r=4$ and use the following code:

Table[Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6], {a, 1, 100, 1}]

It will find that $a=24$ gives $b=70$, so I want to see only that solution.

Maybe I can use the If[] function?

equation-solving programming code-request system code

share|improve this question

asked 8 hours ago

JanJan

25022 silver badges1111 bronze badges

$endgroup$

b==(-12 + 

  Sqrt[3] Sqrt[

   3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 

    4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

Table[Sqrt[a (1 + a) (1 + 2 a)]/Sqrt[6], {a, 1, 100, 1}]

share|improve this question

asked 8 hours ago

JanJan

25022 silver badges1111 bronze badges

share|improve this question

asked 8 hours ago

JanJan

25022 silver badges1111 bronze badges

asked 8 hours ago

JanJan

25022 silver badges1111 bronze badges

asked 8 hours ago

JanJan

25022 silver badges1111 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

You can use Solve. Your equation:

eqn[r_] := b == (-12 + Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

Using Solve (with $r=4$):

Solve[eqn[4] && 1 < a < 100, {a, b}, Integers]

{{a -> 24, b -> 70}}

Using Solve (with $r=11$):

Solve[eqn[11] && 1 < a < 100, {a, b}, Integers]

{{a -> 25, b -> 73}}

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

90.6k33 gold badges120120 silver badges232232 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

You might try IntegerQ

Table[b = (-12 +Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2+ r) +4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r)) /. r -> 4;

If[IntegerQ[b], {b, a}, Nothing], {a, 2, 50}]

(*{{70, 24}}*)

share|improve this answer

answered 7 hours ago

Ulrich NeumannUlrich Neumann

14.3k77 silver badges2323 bronze badges

$endgroup$

add a comment
 | 

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4

$begingroup$

You can use Solve. Your equation:

eqn[r_] := b == (-12 + Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

Using Solve (with $r=4$):

Solve[eqn[4] && 1 < a < 100, {a, b}, Integers]

{{a -> 24, b -> 70}}

Using Solve (with $r=11$):

Solve[eqn[11] && 1 < a < 100, {a, b}, Integers]

{{a -> 25, b -> 73}}

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

90.6k33 gold badges120120 silver badges232232 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

You can use Solve. Your equation:

eqn[r_] := b == (-12 + Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

Using Solve (with $r=4$):

Solve[eqn[4] && 1 < a < 100, {a, b}, Integers]

{{a -> 24, b -> 70}}

Using Solve (with $r=11$):

Solve[eqn[11] && 1 < a < 100, {a, b}, Integers]

{{a -> 25, b -> 73}}

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

90.6k33 gold badges120120 silver badges232232 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You can use Solve. Your equation:

eqn[r_] := b == (-12 + Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

Using Solve (with $r=4$):

Solve[eqn[4] && 1 < a < 100, {a, b}, Integers]

{{a -> 24, b -> 70}}

Using Solve (with $r=11$):

Solve[eqn[11] && 1 < a < 100, {a, b}, Integers]

{{a -> 25, b -> 73}}

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

90.6k33 gold badges120120 silver badges232232 bronze badges

$endgroup$

eqn[r_] := b == (-12 + Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2 + r) + 4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r))

Solve[eqn[4] && 1 < a < 100, {a, b}, Integers]

Solve[eqn[11] && 1 < a < 100, {a, b}, Integers]

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

90.6k33 gold badges120120 silver badges232232 bronze badges

answered 7 hours ago

Carl WollCarl Woll

90.6k33 gold badges120120 silver badges232232 bronze badges

answered 7 hours ago

Carl WollCarl Woll

90.6k33 gold badges120120 silver badges232232 bronze badges

2

$begingroup$

You might try IntegerQ

Table[b = (-12 +Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2+ r) +4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r)) /. r -> 4;

If[IntegerQ[b], {b, a}, Nothing], {a, 2, 50}]

(*{{70, 24}}*)

share|improve this answer

answered 7 hours ago

Ulrich NeumannUlrich Neumann

14.3k77 silver badges2323 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

You might try IntegerQ

Table[b = (-12 +Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2+ r) +4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r)) /. r -> 4;

If[IntegerQ[b], {b, a}, Nothing], {a, 2, 50}]

(*{{70, 24}}*)

share|improve this answer

answered 7 hours ago

Ulrich NeumannUlrich Neumann

14.3k77 silver badges2323 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You might try IntegerQ

Table[b = (-12 +Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2+ r) +4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r)) /. r -> 4;

If[IntegerQ[b], {b, a}, Nothing], {a, 2, 50}]

(*{{70, 24}}*)

share|improve this answer

answered 7 hours ago

Ulrich NeumannUlrich Neumann

14.3k77 silver badges2323 bronze badges

$endgroup$

Table[b = (-12 +Sqrt[3] Sqrt[3 (-4 + r)^2 + 12 a^2 (-2 + r) - 4 a (-5 + r) (-2+ r) +4 a^3 (-2 + r)^2] + 3 r)/(6 (-2 + r)) /. r -> 4;

If[IntegerQ[b], {b, a}, Nothing], {a, 2, 50}]

(*{{70, 24}}*)

share|improve this answer

answered 7 hours ago

Ulrich NeumannUlrich Neumann

14.3k77 silver badges2323 bronze badges

answered 7 hours ago

Ulrich NeumannUlrich Neumann

14.3k77 silver badges2323 bronze badges

answered 7 hours ago

Ulrich NeumannUlrich Neumann

14.3k77 silver badges2323 bronze badges