Is this extension of Goldbach's conjecture obviously false? Goldbach's conjecture is: 

Every even integer greater than $2$ can be expressed as the sum of two primes.

Extension of Goldbach's conjecture is:

Every number from $p\mathbb{Z}$ greater than $p$ can be expressed as the sum of $p$ primes

where $p\mathbb{Z}=\lbrace 0, \pm p, \pm 2p, \pm3p,\dots\rbrace$.
You can find a counterexamples?
For example if $p=5$:


*

*$10=2+2+2+2+2$

*$15=3+3+3+3+3$

*$20=2+3+3+5+7$

*...

*$100=47+2+17+17+17$


Who can make this programming?
 A: Suppose Goldbach is true.  Let n be your candidate number.  Then either n is even or odd.  If even, then it is the sum of 2 primes.  If odd then n - 3 is even so n is the sum of 3 primes.  In no case would you need more than 3 primes.
A: First of all we will assume that the Goldbach's conjecture is true, so given an integer  $n=kp$ with $k\geq 4$ ($k=3,2$ are easy), we will prove that $n$ is the sum of $p$ primes.
If $p=2,$ it's the Goldbach's conjecture and it's true, so we can assume that $p>2$.
Case 1 If $n$ is even, let's write 
$$n=\underbrace{2+\cdots+2}_{p-2}+(n-2(p-2)),$$
and because $(n-2(p-2))=(k-2)p+4\geq4$ and it's even, then it can expressed as the sum of two primes. Hence, $n$ is the sum of $p$ primes.
Case 2 If $n$ is odd, let's write 
$$n=3+\underbrace{2+\cdots+2}_{p-3}+(n-2(p-3)-3),$$
and because $(n-2(p-3)-3)=(k-2)p+3\geq4$ and it's even, then it can expressed as the sum of two primes. Hence, $n$ is the sum of $p$ primes.
So we proved the following statement:
$$\text{Goldbach's conjecture} \implies \text{Extension of Goldbach's conjecture}$$
But Goldbach's conjecture is a particular case of the extension of Goldbach's conjecture (when $p=2$), hence:
$$\text{Goldbach's conjecture} \iff \text{Extension of Goldbach's conjecture}$$
Conclusion The extension of Goldbach's conjecture as you define it is equivalent to the Goldbach's conjecture itself, and hence no one will ever find a counterexample or prove it unless (s)he solves Goldbach's conjecture.
