Hello
I have some problems finding those n's(where n>0 ,integer).We are working with C[x].
I know that x2+1 's roots must be the roots of xn+1 too.So xn +1 =(x-i)*(x+i)*h for some h∈C[x]
n=1 is not while n=2 is obviously good.I tried a few n in wolfram alpha and got that n=2+4k where k is non negative integer.Maybe i could use induction to prove it,but i don't know if an easier way exists.
Anyway,i will try the induction.
For n=0,it works.
Let's say that it works for some n=2+4k,so exists an f∈C[x] that xn+1=(x2+1)*f.
We have to prove that exists a g∈C[x] that xn+1 +1=(x2+1)*g
Using xn+1=(x2+1)*f,we get that xn+1+1=(x2+1)*f*x -x +1.
Now we have to take a look at f*x -x +1.
If deg(f)=0,then f=1 is a must.If deg(f)>0,then f*x -x +1 can be substituted with a product,using canonical form of it(becouse the fundamental theorem of algebra).SO we get xn+1+1=(x2+1)*e,where e=(x-f1)(x-f2)...(x-fl) [f1,f2... are the roots of f*x -x +1].If we say e=g,the statement if proved.
I have some problems finding those n's(where n>0 ,integer).We are working with C[x].
I know that x2+1 's roots must be the roots of xn+1 too.So xn +1 =(x-i)*(x+i)*h for some h∈C[x]
n=1 is not while n=2 is obviously good.I tried a few n in wolfram alpha and got that n=2+4k where k is non negative integer.Maybe i could use induction to prove it,but i don't know if an easier way exists.
Anyway,i will try the induction.
For n=0,it works.
Let's say that it works for some n=2+4k,so exists an f∈C[x] that xn+1=(x2+1)*f.
We have to prove that exists a g∈C[x] that xn+1 +1=(x2+1)*g
Using xn+1=(x2+1)*f,we get that xn+1+1=(x2+1)*f*x -x +1.
Now we have to take a look at f*x -x +1.
If deg(f)=0,then f=1 is a must.If deg(f)>0,then f*x -x +1 can be substituted with a product,using canonical form of it(becouse the fundamental theorem of algebra).SO we get xn+1+1=(x2+1)*e,where e=(x-f1)(x-f2)...(x-fl) [f1,f2... are the roots of f*x -x +1].If we say e=g,the statement if proved.