My discovery about mathematics

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I give an answer to the 12th problem of Hilbert by solutions of Schröder equation.

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Let H(a) be the set of the complex functions which are holomorphic at a∈C.
Let's use the next theorem.
Theorem 1. If g(z)=(z/2)+∑_n∈N-{1} a_nzⁿ ∈H(0),there is a function t(z) for which
t(z)∈H(0) and ∃r((r∈R-{0})∧((|z|≤|r|)→(t(z)/2=t(g(z)))) and t´(0)=1.
Let f(x)=2x+∑_n∈N-{1} c_nxⁿ be the inverse function of g(z). Put x=g(z). f(x)=z.
So 2t(x)=t(f(x)) for t(z) in theorem 1. Let x(t) be the inverse function of t(x).
Theorem 2. If f(x) is an integral function, you can prolong x(t) as an integral function for
which x(2t)=f(x(t)) .
Proof. ∃r"((r"∈R-{0})∧((|2t|≤|r"|)→(x(2t)=f(x(t)))). x(2t) is holomolphic on |2t|≤r".
Define x(2t)=f(x(t)) on |t|≤r". x(2t) is holomorphic on |2t|≤2r"
and x(2t)=f(x(t)) on |2t|≤2r".
Repeating this, you get an integral function x(t) for which x(2t)=f(x(t)) on t∈C.♦
Theorem 3. x(t+c)=x(t) (t∈C) if x(c)=0 and c≠0 and
Proof. Put K={ t |(t∈C)∧(x(t+c)=x(t))}. If K isn't a discrete set, x(t+c)=x(t) for ∀t∈C.
Put F(t,u)=x(t)x(u)/(x(t+u)-x(t)-x(u)).F(t,t)=x(t)²/(x(2t)-2x(t))=1/(∑_n≥2c_nx(t)^n-2).
F(t,t)→1/c₂≠0 (t→c). F(t,u)→1/c₂ (t→c, u=t→c).
c∈K. Put G(t)=x(t)x(c)/(x(t+c)-x(t)-x(c)). t∈C-K ⇒ G(t)=0.
If K is a discrete set, ∃d((d∈R-{0})∧((|t-c|≤|d|)∧(t≠c)→(G(t)=0)). G(t)=0 on t∈C--K.
F(t,u)=G(t)→0 (t→c,u=c→c). When t=c→c and u=c→c , F(t,u) isn't decided for 1/c₂≠0.
But x(t)=∑_n∈N e_n(t-c)ⁿ.
This is contradiction.♦
When f(x)=p(x^N-x-q,f´(x)=Nx^N-1-1=0⇒ x/N^N-1=1. x∈Q.
Let K/Q be Galois extention. ∃p(x)((p(x)∈Q[x])∧(p(0)≠0)∧((p(u)=0)→(K=Q(u)).
p(x) has no double root. You can set up that f(x)=2p(x)x/p(0). Let w be a period of x(t).
x(w)=x(0+w)=x(0)=0. ∃n((x(w/2^n-1)=0)∧(x(w/2ⁿ)≠0)).
0=x(w/2^n-1)=2p(x(w/2ⁿ))x(w/2ⁿ)/p(0). p(x(w/2ⁿ))=0. K=Q(x(w/2ⁿ)).
This is an answer to the 12th problem of Hilbert by x(t)=x(t ; c₂,…,c_m).

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Are the x(t ; c₂,…,c_m) 's known functions like x(t ; 1)=e^t-1?