<s>[INST] Primality Criterion for Specific Class of Proth Numbers Is this proof acceptable ?
Theorem :
Let  $N = k\cdot 2^n+1$  with  $n>1$ , $k<2^n$  ,  $3 \mid k $ , and
$\begin{cases}
 k \equiv 3 \pmod {30} , & \text{with }n \equiv 1,2 \pmod 4 \\
 k \equiv 9 \pmod {30} , & \text{with }n \equiv 2,3 \pmod 4 \\
 k \equiv 21 \pmod {30} , & \text{with }n \equiv 0,1 \pmod 4 \\
 k \equiv 27 \pmod {30} , & \text{with }n \equiv 0,3 \pmod 4
 \end{cases}$ 
, thus
$N$ is prime iff $5^{\frac{N-1}{2}} \equiv -1 \pmod N$
Proof :
Necessity : $\textit{If}~ N ~\textit{is  prime then} ~ 5^{\frac{N-1}{2}} \equiv -1 \pmod N $
Let $N$ be a prime , then by  Euler criterion :
$5^{\frac{N-1}{2}} \equiv \left(\frac{5}{N}\right) \pmod N$ 
If $N$ is  prime then $N \equiv 2,3 \pmod 5 $ and therefore : $\left(\frac{N}{5}\right)=-1$ . 
Since $N \equiv 1 \pmod 4 $ according to the law of quadratic reciprocity it follows that :  $\left(\frac{5}{N}\right)=-1$ .
Hence , $5^{\frac{N-1}{2}} \equiv -1 \pmod N $ . 
Sufficiency : $\textit{If}~ 5^{\frac{N-1}{2}} \equiv -1 \pmod N~ \textit{then}~ N~ \textit{is  prime}$ 
If $5^{\frac{N-1}{2}} \equiv -1 \pmod N$ then by Proth's theorem $N$ is  prime .
 [/INST] You just took two (true) theorems and emerged them into one, and based your proof, just by assuming both theorems are true.
For me it is unacceptable.
Proving first part is just too easy, but it is always nice to also show why you used or took as granted the given formulas. Not many people heard about Legendre symbol.
Proof of second part is just in the (topic of) question itself.
It didn't take 8 lines for Euler or Proth to prove their theorems.
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