Mixing
Could someone explain how to solve this problem?
-2
$begingroup$
Let $$f(x)=x^2-3$$
For how many integer values of x is $f(f(f(x)))$ divisible by x?
integers
edited Jan 30 at 16:17
Thomas Andrews
131k12147298
asked Jan 30 at 16:13
kingking
12
$endgroup$
add a comment |
-2
$begingroup$
Let $$f(x)=x^2-3$$
For how many integer values of x is $f(f(f(x)))$ divisible by x?
integers
edited Jan 30 at 16:17
Thomas Andrews
131k12147298
asked Jan 30 at 16:13
kingking
12
$endgroup$
$begingroup$
I don't know how to solve it, but I know what I would do to try to find a solution. The first thing would be to insert $x=0,1,2,3$ or $4$ and see what pops out. You should try that. And please let us know the results.
$endgroup$
– Arthur
Jan 30 at 16:20
$begingroup$
In general, if $h$ is an integer polynomial and $x$ is an integer, then $x$ is a divisor of $h(x)-h(0).$ So $x$ is a divisor of $h(x)$ if and only if $x$ is a divisor of $h(0).$
$endgroup$
– Thomas Andrews
Jan 30 at 17:15
$begingroup$
Could you give me more hints? do you mean I should solve for h(0)?
$endgroup$
– king
Jan 31 at 1:13
add a comment |
-2
-2
-2
$begingroup$
Let $$f(x)=x^2-3$$
For how many integer values of x is $f(f(f(x)))$ divisible by x?
integers
edited Jan 30 at 16:17
Thomas Andrews
131k12147298
asked Jan 30 at 16:13
kingking
12
$endgroup$
Let $$f(x)=x^2-3$$
For how many integer values of x is $f(f(f(x)))$ divisible by x?
integers
integers
edited Jan 30 at 16:17
Thomas Andrews
131k12147298
asked Jan 30 at 16:13
kingking
12
edited Jan 30 at 16:17
Thomas Andrews
131k12147298
asked Jan 30 at 16:13
kingking
12
edited Jan 30 at 16:17
Thomas Andrews
131k12147298
edited Jan 30 at 16:17
Thomas Andrews
131k12147298
edited Jan 30 at 16:17
Thomas Andrews
131k12147298
131k12147298
asked Jan 30 at 16:13
kingking
12
asked Jan 30 at 16:13
kingking
12
asked Jan 30 at 16:13
kingking
12
12
$begingroup$
I don't know how to solve it, but I know what I would do to try to find a solution. The first thing would be to insert $x=0,1,2,3$ or $4$ and see what pops out. You should try that. And please let us know the results.
$endgroup$
– Arthur
Jan 30 at 16:20
$begingroup$
In general, if $h$ is an integer polynomial and $x$ is an integer, then $x$ is a divisor of $h(x)-h(0).$ So $x$ is a divisor of $h(x)$ if and only if $x$ is a divisor of $h(0).$
$endgroup$
– Thomas Andrews
Jan 30 at 17:15
$begingroup$
Could you give me more hints? do you mean I should solve for h(0)?
$endgroup$
– king
Jan 31 at 1:13
add a comment |
$begingroup$
I don't know how to solve it, but I know what I would do to try to find a solution. The first thing would be to insert $x=0,1,2,3$ or $4$ and see what pops out. You should try that. And please let us know the results.
$endgroup$
– Arthur
Jan 30 at 16:20
$begingroup$
In general, if $h$ is an integer polynomial and $x$ is an integer, then $x$ is a divisor of $h(x)-h(0).$ So $x$ is a divisor of $h(x)$ if and only if $x$ is a divisor of $h(0).$
$endgroup$
– Thomas Andrews
Jan 30 at 17:15
$begingroup$
Could you give me more hints? do you mean I should solve for h(0)?
$endgroup$
– king
Jan 31 at 1:13
$begingroup$
I don't know how to solve it, but I know what I would do to try to find a solution. The first thing would be to insert $x=0,1,2,3$ or $4$ and see what pops out. You should try that. And please let us know the results.
$endgroup$
– Arthur
Jan 30 at 16:20
$begingroup$
I don't know how to solve it, but I know what I would do to try to find a solution. The first thing would be to insert $x=0,1,2,3$ or $4$ and see what pops out. You should try that. And please let us know the results.
$endgroup$
– Arthur
Jan 30 at 16:20
$begingroup$
In general, if $h$ is an integer polynomial and $x$ is an integer, then $x$ is a divisor of $h(x)-h(0).$ So $x$ is a divisor of $h(x)$ if and only if $x$ is a divisor of $h(0).$
$endgroup$
– Thomas Andrews
Jan 30 at 17:15
$begingroup$
In general, if $h$ is an integer polynomial and $x$ is an integer, then $x$ is a divisor of $h(x)-h(0).$ So $x$ is a divisor of $h(x)$ if and only if $x$ is a divisor of $h(0).$
$endgroup$
– Thomas Andrews
Jan 30 at 17:15
$begingroup$
Could you give me more hints? do you mean I should solve for h(0)?
$endgroup$
– king
Jan 31 at 1:13
$begingroup$
Could you give me more hints? do you mean I should solve for h(0)?
$endgroup$
– king
Jan 31 at 1:13
add a comment |
1 Answer
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oldest
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$begingroup$
Hint: $$f(f(x))=(x^2-3)^2-3$$ and now what is $$f(f(f(x))$$?
answered Jan 30 at 16:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
The f(f(f(x)) becomes very complicated, I don't know the next step to solve this problem
$endgroup$
– king
Jan 31 at 1:06
add a comment |
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1 Answer
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1 Answer
1
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1
$begingroup$
Hint: $$f(f(x))=(x^2-3)^2-3$$ and now what is $$f(f(f(x))$$?
answered Jan 30 at 16:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
The f(f(f(x)) becomes very complicated, I don't know the next step to solve this problem
$endgroup$
– king
Jan 31 at 1:06
add a comment |
1
$begingroup$
Hint: $$f(f(x))=(x^2-3)^2-3$$ and now what is $$f(f(f(x))$$?
answered Jan 30 at 16:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
The f(f(f(x)) becomes very complicated, I don't know the next step to solve this problem
$endgroup$
– king
Jan 31 at 1:06
add a comment |
1
1
1
$begingroup$
Hint: $$f(f(x))=(x^2-3)^2-3$$ and now what is $$f(f(f(x))$$?
answered Jan 30 at 16:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
Hint: $$f(f(x))=(x^2-3)^2-3$$ and now what is $$f(f(f(x))$$?
answered Jan 30 at 16:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 30 at 16:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 30 at 16:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 30 at 16:21
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
$begingroup$
The f(f(f(x)) becomes very complicated, I don't know the next step to solve this problem
$endgroup$
– king
Jan 31 at 1:06
add a comment |
$begingroup$
The f(f(f(x)) becomes very complicated, I don't know the next step to solve this problem
$endgroup$
– king
Jan 31 at 1:06
$begingroup$
The f(f(f(x)) becomes very complicated, I don't know the next step to solve this problem
$endgroup$
– king
Jan 31 at 1:06
$begingroup$
The f(f(f(x)) becomes very complicated, I don't know the next step to solve this problem
$endgroup$
– king
Jan 31 at 1:06
add a comment |
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$begingroup$
I don't know how to solve it, but I know what I would do to try to find a solution. The first thing would be to insert $x=0,1,2,3$ or $4$ and see what pops out. You should try that. And please let us know the results.
$endgroup$
– Arthur
Jan 30 at 16:20
$begingroup$
In general, if $h$ is an integer polynomial and $x$ is an integer, then $x$ is a divisor of $h(x)-h(0).$ So $x$ is a divisor of $h(x)$ if and only if $x$ is a divisor of $h(0).$
$endgroup$
– Thomas Andrews
Jan 30 at 17:15
$begingroup$
Could you give me more hints? do you mean I should solve for h(0)?
$endgroup$
– king
Jan 31 at 1:13